8. Quadratic Functions, Graphs and Equations
This Chapter discusses quadratic functions and their graphs. The quadratic is a common algebraic expression and its properties should be familiar to you. Quadratics arise in maths, business, engineering and science.
The following topics are discussed
- Definition of a quadratic function.
- Completing the square
- The turning point
- Maximum and minimum values
- Graphs of quadratic functions
- The Discriminant
- The Quadratic formula
- Higher order equations.
Do you need this chapter?
Here are some questions on the topics in this chapter to consider. If you can do these you may skip this chapter.
Quiz: Quadratic functions
Here are some questions that illustrate the material discussed in this chapter.
1. Sketch the graph of [latex]y=2x^2[/latex].
Answer
2. Complete the square of [latex]2x^2+3x+4[/latex].
Answer
[latex]2\left( x + \dfrac{3}{4} \right) ^2 + \dfrac{23}{8}[/latex]
3. Sketch the graph of [latex]y=\left(x-1\right)^2[/latex].
Answer
4. Sketch the graph of [latex]y=x^2-5x+6[/latex].
Answer
5. The profit from selling an article in dollars [latex]p[/latex] is related to the number of articles produced [latex]n[/latex] by the formula
\begin{align*}
p&=-n^2+10n.
\end{align*}
How many articles should be made to maximize the profit?
Answer
Five articles.
6. How many real solutions does the quadratic equation [latex]7x^2-2x-1=0[/latex] have?
Answer
Two solutions.
7. What is the turning point of the graph of the function [latex]y=3x^2-4x+1[/latex]?
Answer
Turning point at [latex]\left(\dfrac{2}{3},-\dfrac{1}{3} \right)[/latex]
8. If [latex]m^2-8m=0[/latex], [latex]m=[/latex]?
a) [latex]1[/latex] or [latex]8[/latex]
b) [latex]0[/latex] or [latex]4[/latex]
c) [latex]0[/latex] or [latex]8[/latex]
d) [latex]1[/latex] or [latex]4[/latex]
e) None of the above.
Answer
c)
9. If [latex]u^2 - u -30 =0[/latex] which of the following is NOT true?
a) [latex]u=6[/latex] is a solution
b) [latex]u=10[/latex] is a solution
c) there are two possible solutions
d) [latex]u=-5[/latex] is a solution
e) [latex]0[/latex] is not a solution.
Answer
b)
10. Solve for [latex]m[/latex]: [latex]m^{2}-8m +15=0[/latex]
a) [latex]m=-5[/latex] and [latex]m=3[/latex]
b) [latex]m=-5[/latex] and [latex]m=-3[/latex]
c) [latex]m=-4[/latex] and [latex]m=2[/latex]
d) [latex]m=-2[/latex] and [latex]m=-4[/latex]
e) None of the above.
Answer
e)
11. How many times does the graph of the parabola with equation [latex]y=918x^2+93x-1[/latex] cut the [latex]x-[/latex]axis? There is no need to solve the equation.
a) It doesn't cut the [latex]x-[/latex] axis
b) [latex]3[/latex] times
c) [latex]2[/latex] times
d) once
e) None of the above.
Answer
c)
12. Solve the equation [latex]x^4-5x^2+4=0[/latex].
Answer
[latex]x=-2, \ x=-1,\ x=1,\ x=2[/latex]
13. Solve the equation [latex]3e^{2x}-7e^x+2=0[/latex].
Answer
[latex]x=\ln\left(1/3 \right)[/latex] and [latex]x=\ln \left( 2 \right)[/latex]
Quadratic functions
A quadratic function [latex]f\left(x\right)[/latex], has the form
\begin{align*}
f\left(x\right)&=ax^2+bx+c
\end{align*}
where [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are real constants and [latex]a\neq 0[/latex].
For example
\begin{align*}
f\left(x\right)&=x^2\\
f\left(x\right)&=3x^2\\
f\left(x\right)&=2x^2+1\\
f\left(x\right)&=2x^2+3x\\
f\left(x\right)&=5x^2+2x-7\\
f\left(x\right)&=\left(x-2\right)^2\\
f\left(x\right)&=\left(x-2\right)^2-5\\
\end{align*}
are all quadratic functions.
Plotting graphs of quadratic functions
The graph of [latex]y=x^2[/latex] may be obtained by plotting coordinates. The following table gives some points on the graph.
| [latex]x=[/latex] | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| [latex]y=x^2[/latex] | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
These points and the corresponding graph are shown in Fig. 8.1,
Note that the characteristic shape of the graph of [latex]y=x^2[/latex] is a parabola facing up (or concave up). It has one minimum value of [latex]0[/latex] at [latex]x=0[/latex].
What about the graph of [latex]y=-x^2\,[/latex]? The corresponding points are shown in the table below.
| [latex]x=[/latex] | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| [latex]y=-x^2[/latex] | -9 | -4 | -1 | -0 | -1 | -4 | -9 |
The graph is shown in Fig. 8.2.
Note that it is concave down and has a maximum of [latex]0[/latex] at [latex]x=0[/latex].
The graphs of all quadratic functions are either concave up, when [latex]a[/latex], the coefficient of [latex]x^2[/latex], is greater than zero, or concave down when [latex]a[/latex] is less than zero. The minimum or maximum value of the function occurs at the turning point. The graphs are symmetrical about a vertical line through the turning point. We will discuss this in detail in later sections.
The location and general shape of the graph are determined by the values of the constants [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex]. Click here for an interactive quadratic graph. Use the sliders to adjust the values of [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] and see the effect on the graph of [latex]y=ax^2+bx+c[/latex].
Completing the square
Recall the binomial rules:
\begin{align*}
a^2+2ab+b^2&=\left(a+b \right)^2\\
a^2-2ab+b^2&=\left(a-b \right)^2.\\
\end{align*}
You should know these rules by heart. We will use them frequently when completing the square.
Completing the square is a useful technique for establishing the turning point and the maximum or minimum value of a quadratic function. The steps in completing the square are:
- Divide the quadratic function [latex]ax^2+bx+c[/latex] by [latex]a[/latex] to change it to the form
\begin{align*}
a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right].
\end{align*}
Of course, this is unnecessary if [latex]a=1[/latex]. - Take half of the [latex]x[/latex]-coefficient, square it, and add and subtract it to the function to get
\begin{align*}
a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right]&=a\left[x^2+\frac{b}{a}x+\underbrace{\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2}_\text{these terms equal zero}+\frac{c}{a}\right]\\
\end{align*}
Note that this, in effect, does nothing as you are adding zero. However, it does allow us to rewrite the expression in a useful form in the following step. - Rearrange the formula using the binomial rule to get
\begin{align*}
a\left[x^2+\frac{b}{a}x+\frac{c}{a}\right]&=a\left[ \underbrace{x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2}_\text{use binomial rule on these terms}-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\right]\\
&=a\left[ \left( x+\frac{b}{2a} \right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\right]\\
\end{align*} - Rearrange the resulting expression. This usually involves multiplication by [latex]a[/latex] and simplification of the resulting fraction.
The procedure looks complicated but is straight forward in practice as shown in the following examples.
Examples: Completing the square
1. Complete the square on [latex]x^2+4x+4[/latex].
Solution:
In this case [latex]a=1[/latex] and there is no need to divide the expression. The [latex]x[/latex] coefficient is [latex]4[/latex]. We take half of this and add and subtract its square to the function to get:
\begin{align*}
x^2+4x+4&= x^2 +4x+\underbrace{\left( \frac{4}{2}\right)^2- \left( \frac{4}{2} \right)^2}_\text{adds 0} +4 \\
&=x^2 + 4x+2^2- 2^2 +4 \\
&=\underbrace{x^2 +4x+4}_\text{apply binomial rule}- 4 +4 \\
&= \left(x +2 \right)^2- 4+4, \text{using the binomial rules}\\
&=\left(x+2 \right)^2.
\end{align*}
2. Complete the square on [latex]x^2-6x+9[/latex].
Solution:
In this case [latex]a=1[/latex] and there is no need to divide the expression. The [latex]x[/latex] coefficient is [latex]-6[/latex]. We take half of this and add and subtract its square to the function to get:
\begin{align*}
x^2-6x+9&= x^2-6x+\underbrace{\left(- \frac{6}{2}\right)^2- \left(- \frac{6}{2} \right)^2}_\text{adds 0} +9 \\
&=x^2-6x+\left(-3\right)^2- \left(-3\right)^2 +9 \\
&=\underbrace{x^2-6x+9}_\text{apply binomial rule}- 9 +9 \\
&= \left(x-3 \right)^2- 9+9\\
&=\left(x-3 \right)^2.
\end{align*}
3. Complete the square on [latex]x^2+4x+5[/latex].
Solution:
In this case [latex]a=1[/latex] and there is no need to divide the expression. The [latex]x[/latex] coefficient is [latex]4[/latex]. We take half of this and add and subtract its square to the function to get
\begin{align*}
x^2+4x+5&=x^2+4x+\left( \frac{4}{2} \right)^2-\left( \frac{4}{2} \right)^2+5\\
&=x^2+4x+2^2-2^2+5\\
&=x^2+4x+4-4+5\\
&=\left(x+2 \right)^2-4+5\\
&=\left(x+2 \right)^2+1.\\
\end{align*}
Note the use of [latex]\left(a+b \right)^2=a^2+2ab+b^2[/latex] on the first three terms in line 3 of the solution.
4. Complete the square on [latex]x^2-6x+15[/latex].
Solution:
Again [latex]a=1[/latex] and there is no need to divide the expression. The [latex]x[/latex] coefficient is [latex]-6[/latex]. We take half of this and add and subtract its square to the function to get
\begin{align*}
x^2-6x+15&=x^2-6x+\left( -\frac{6}{2} \right)^2-\left(- \frac{6}{2} \right)^2+15\\
&=x^2-6x+\left( -3 \right)^2-\left(- 3 \right)^2+15\\
&=x^2-6x+9-9+15\\
&=\left(x-3 \right)^2-9+15\\
&=\left(x-3 \right)^2+6.\\
\end{align*}
Note the use of [latex]\left(a-b \right)^2=a^2-2ab+b^2[/latex] on the first three terms in line 3 of the solution.
5. Complete the square on [latex]2x^2+3x+4[/latex].
Solution:
In this case [latex]a=2[/latex] so we divide the expression by [latex]2[/latex] and complete the square on the resulting expression. We have
\begin{align*}
2x^2+3x+4&=2\left(x^2 + \frac{3}{2}x+\frac{4}{2} \right)\\
&=2\left[x^2 + \frac{3}{2}x+2 \right].\\
\end{align*}
Now we complete the square on the expression in square brackets to get:
\begin{align*}
2x^2+3x+4&=2\left[ x^2 + \frac{3}{2}x+2 \right]\\
&=2\left[ x^2 + \frac{3}{2}x+\left( \frac{3}{4}\right)^2- \left( \frac{3}{4} \right)^2 +2\right]\\
&=2\left[ x^2 + \frac{3}{2}x+\frac{9}{16} - \frac{9}{16} +2\right]\\
&=2\left[ \left(x + \frac{3}{4} \right)^2- \frac{9}{16} +2\right]\\
&=2\left[ \left(x + \frac{3}{4} \right)^2- \frac{9}{16} +\frac{32}{16}\right]\\
&=2\left[ \left(x + \frac{3}{4} \right)^2+\frac{23}{16} \right]\\
&=2 \left(x + \frac{3}{4} \right)^2+\frac{23}{8} .
\end{align*}
6. Complete the square on [latex]3x^2-2x-1[/latex].
Solution:
Using the same method as in the previous example, we take out a factor of 3 and then complete the square. We have:
\begin{align*}
3x^2-2x-1&=3\left[ x^2-\frac{2}{3}x-\frac{1}{3} \right] \\
&=3\left[ x^2-\frac{2}{3}x+\left( \frac{1}{3}\right)^2- \left( \frac{1}{3} \right)^2 -\frac{1}{3}\right] \\
&=3\left[ x^2-\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}-\frac{1}{3}\right] \\
&=3\left[ \left(x-\frac{1}{3} \right)^2- \frac{1}{9} -\frac{1}{3}\right]) \\
&=3\left[ \left(x-\frac{1}{3} \right)^2- \frac{1}{9} -\frac{3}{9}\right] \\
&=3\left[ \left(x-\frac{1}{3} \right)^2- \frac{4}{9} \right]\\
&=3\left(x-\frac{1}{3} \right)^2- \frac{4}{3}.
\end{align*}
Here are some exercises to try.
Exercises: Completing the square
Complete the square on the following.
1. [latex]x^2+2x+1[/latex]
2. [latex]x^2-8x+16[/latex]
3. [latex]x^2+6x-15[/latex]
4. [latex]x^2+5x-6[/latex]
5. [latex]3x^2-5x+7[/latex]
Answers
1. [latex]\left( x+1 \right)^2[/latex]
2. [latex]\left( x-4 \right)^2[/latex]
3. [latex]\left( x+3 \right)^2-24[/latex]
4. [latex]\left( x+\dfrac{5}{2} \right)^2-\dfrac{49}{4}[/latex]
5. [latex]3\left( x- \dfrac{5}{6} \right)^2+\dfrac{59}{12}[/latex]
Maximum, minimum values and the turning point
We have seen that the maximum or minimum of a quadratic function occurs at the turning point. Completing the square allows us to find the turning point and hence the maximum or minimum value.
Completing the square on the general form of a quadratic we obtain
\begin{align*}
ax^2+bx+c &= a\left[x^2+\frac{b}{a}x+\frac{c}{a} \right]\\
&= a\left[x^2+\frac{b}{a}x+\left( \frac{b}{2a} \right)^2-\left( \frac{b}{2a} \right)^2+\frac{c}{a} \right]\\
&= a\left[x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a} \right]\\
&= a\left[ \left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a^2}+\frac{c}{a} \right]\\
&= a\left[\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2} \right]\\
&= a\left[\left(x+\frac{b}{2a} \right)^2-\frac{b^2-4ac}{4a^2} \right]\\
&= a \left(x+\frac{b}{2a} \right)^2-\frac{b^2-4ac}{4a}.
\end{align*}
Now suppose that [latex]a>0[/latex]. Then the first term [latex]a \left(x+\dfrac{b}{2a} \right)^2[/latex] will be positive. Moreover, it will increase without limit as [latex]x \rightarrow \infty[/latex] and [latex]x \rightarrow -\infty[/latex]. Hence there can be no maximum in this case, and the graph will be concave up. The quadratic function will have a minimum at the turning point. The minimum value of the quadratic will occur when [latex]a \left(x+\dfrac{b}{2a} \right)^2=0[/latex], that is at [latex]x=-\dfrac{b}{2a}[/latex], and the value of the minimum will be [latex]-\dfrac{b^2-4ac}{4a}[/latex].
On the other hand, if [latex]a \lt 0[/latex], the term [latex]a \left(x+\dfrac{b}{2a} \right)^2[/latex] will approach [latex]-\infty[/latex] as [latex]x \rightarrow \infty[/latex] and [latex]x \rightarrow -\infty[/latex]. The graph will be concave down. We therefore have a maximum at the turning point and this will again occur when [latex]x=-\dfrac{b}{2a}[/latex], and the value of the minimum will be [latex]-\dfrac{b^2-4ac}{4a}[/latex].
To summarise, if [latex]a>0[/latex], the graph will be concave up and have a minimum value at the turning point [latex]x=-\dfrac{b}{2a}[/latex]. If [latex]a \lt 0[/latex], the graph will be concave down and have a maximum at the turning point [latex]x=-\dfrac{b}{2a}[/latex].
Examples: Maximum, minimum values and the turning point
1. Find the value and location of the maximum or minimum of the quadratic function [latex]f\left( x \right)=x^2-5x+6[/latex].
There are two approaches to this problem: 1. Use the turning point formula or 2. Complete the square.
Solution 1. Use the turning point formula.
In this case [latex]a=1[/latex], [latex]b=-5[/latex] and [latex]c=6[/latex]. Since [latex]a>0[/latex], the graph is concave up. Hence we are seeking a minimum. We know the turning point occurs at
\begin{align*}
x&= -\frac{b}{2a} \\
&= -\frac{-5}{2} \\
&=\frac{5}{2} .\\
\end{align*}
The minimum value is
\begin{align*}
f\left(\frac{5}{2}\right)&=\left( \frac{5}{2}\right)^2-5\times \frac{5}{2}+6 \\
&=\frac{25}{4}-\frac{25}{2}+6\\
&=\frac{25}{4}-\frac{50}{4}+\frac{24}{4} \\
&=-\frac{1}{4}.
\end{align*}
Hence [latex]f\left( x \right)=x^2-5x+6[/latex] has a minimum value of [latex]-\dfrac{1}{4}[/latex] at [latex]x=\dfrac{5}{2}[/latex].
Solution 2. Complete the square.
We have
\begin{align*}
f\left( x \right)&=x^2-5x+6\\
&=x^2-5x+\left( \frac{5}{2}\right)^2- \left( \frac{5}{2} \right)^2 +6 \\
&= \left(x-\frac{5}{2} \right)^2-\frac{25}{4}+6\\
&= \left(x-\frac{5}{2} \right)^2-\frac{25}{4}+\frac{24}{4}\\
&=\left(x-\frac{5}{2} \right)^2-\frac{1}{4}.\\
\end{align*}
The minimum value occurs when the first bracket is zero. That is at [latex]x=\dfrac{5}{2}[/latex]. When this bracket is zero, [latex]f\left( x \right)=-\dfrac{1}{4}[/latex].
Hence as we found in Solution [latex]1[/latex], [latex]f\left( x \right)=x^2-5x+6[/latex] has a minimum value of [latex]-\dfrac{1}{4}[/latex] at [latex]x=\dfrac{5}{2}[/latex].
2. Find the value and location of the maximum or minimum of the quadratic function [latex]f\left( x \right)=-3x^2-2x+6[/latex].
Solution:
It is possible to use the turning point formula but we will complete the square.
Since [latex]a\lt0[/latex], the graph is concave down and we are seeking a maximum. Completing the square,
\begin{align*}
f\left( x \right)&=-3x^2-2x+6\\
&=-3\left[ x^2 - \frac{2}{-3}x-2 \right] \\
&=-3\left[ x^2+ \frac{2}{3}x+\left( \frac{1}{3}\right)^2- \left( \frac{1}{3} \right)^2 -2 \right] \\
&=-3\left[ \left(x+ \frac{1}{3} \right)^2- \frac{1}{9}-\frac{18}{9} \right] \\
&=-3\left[ \left(x+ \frac{1}{3} \right)^2- \frac{19}{9}\right]\\
&=-3 \left(x+ \frac{1}{3} \right)^2+ \frac{19}{3}.
\end{align*}
As the first term is negative, the maximum is achieved when it is zero. This occurs at [latex]x=-\dfrac{1}{3}[/latex] and the function takes a maximum value of [latex]\dfrac{19}{3}[/latex].
3. Sketch the graph of [latex]y=x^2-2x+9[/latex].
Solution:
Since [latex]a>0[/latex] the graph is concave up and so has a minimum. Completing the square, we have
\begin{align*}
y&=x^2-2x+9\\
&=x^2-2x+1^2-1^1+9\\
&=\left(x-1 \right)^2-1+9\\
&=\left(x-1 \right)^2+8.\\
\end{align*}
Hence the minimum occurs when the first bracket is zero, at [latex]x=1[/latex]. The graph has the value [latex]8[/latex] there. Hence the the turning point is [latex]\left( 1,8 \right)[/latex].
The [latex]y[/latex]-intercept occurs when [latex]x=0[/latex]. Setting [latex]x=0[/latex] gives
\begin{align*}
y&=\left(0-1 \right)^2+8\\
&=9.
\end{align*}
Hence the [latex]y[/latex]-intercept is [latex]\left( 0,9 \right)[/latex]. Noting that the graph is symmetric about a vertical line through the turning point, we can plot the turning point and [latex]y[/latex]-intercept and sketch the graph as shown in Fig. 8.3.
4. The profit [latex]p[/latex], in thousands of dollars, is related to the number [latex]n[/latex], of articles produced by the formula
\begin{align*}
p&=-n^2+10n.
\end{align*}
How many articles should be made to maximize the profit? What is the maximum profit?Solution:
We note that the profit is a quadratic function of [latex]n[/latex]. As the first term is negative, the graph of [latex]p[/latex] is concave down and so will have a maximum. Completing the square:
p&=-n^2+10n\\
&=-\left[n^2-10n\right]\\
&=-\left[n^2-10n +25-25\right]\\
&=-\left[ \left(n-5\right)^2-25\right]\\
&=-\left(n-5\right)^2+25.
\end{align*}
We see that the maximum occurs at [latex]n=5[/latex]. When [latex]n=5[/latex] the profit [latex]p=25[/latex]. Hence, to maximise profit, you should manufacture [latex]5[/latex] articles. The maximum profit is [latex]$25,000[/latex].
The Quadratic formula and the discriminant
The graph shown in Fig. 8.3 for example [latex]3[/latex] above did not touch or cut the [latex]x[/latex]-axis. This is the same as saying that the quadratic function is never zero. Conversely the graph will intersect the [latex]x[/latex]-axis when the quadratic function is zero that is
\begin{align*}
0&=ax^2+bx+c\\
\end{align*}
where [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are constants. We can find the [latex]x[/latex] values for which this is true by completing the square on the right hand side. We have
\begin{align*}
0&=ax^2+bx+c\\
&=a\left[ x^2+\frac{b}{a}x+\left( \frac{b}{2a} \right)^2-\left( \frac{b}{2a} \right)^2+\frac{c}{a} \right] \\
&= \left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2} +\frac{c}{a} \\
&= \left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2} +\frac{4ac}{4a^2}. \\
\end{align*}
This may be rearranged to give
\begin{align*}
\left(x+\frac{b}{2a}\right)^2&=\frac{b^2-4ac}{4a^2}. \\
\end{align*}
Taking the square root of both sides we obtain:
\begin{align*}
x+\frac{b}{2a}&=\pm \sqrt{\frac{b^2-4ac}{4a^2}}. \\
\end{align*}
Another rearrangement yields
\begin{align*}
x &=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. \\
\end{align*}
This formula is called the quadratic formula. It gives the values of [latex]x[/latex] for which [latex]ax^2+bx+c=0[/latex]. The quantity [latex]\Delta=b^2-4ac[/latex] is called the discriminant and determines how many solutions there are. It can be seen from the quadratic formula that if,
- [latex]\Delta>0[/latex], there will be two real solutions and hence two intersections of the graph with the [latex]x[/latex]-axis. These occur at
\begin{align*}
x &=\frac{-b\pm \sqrt{\Delta}}{2a}. \\
\end{align*} - [latex]\Delta=0[/latex], there will be one solution and hence one point where the graph touches the [latex]x[/latex]-axis.
This occurs at
\begin{align*}
x &=-\frac{b}{2a} \\
\end{align*}
and - [latex]\Delta\lt 0[/latex], there will be no real solutions and hence no intersections of the graph with the [latex]x[/latex]-axis because we can't take the square root of a negative number. Note that, in this case, there are complex solutions, but we do not consider them here.
Figure 8.4 illustrates these three cases when the graph is concave up.
Examples: The Discriminant
1. Evaluate the discriminant of the following equations:
\begin{align*}
\quad a)\ x^2-11x+3\qquad b)\ -3x^2+100 \qquad c)\ 5x^2\\
\end{align*}
Solution a)
We have [latex]a=1[/latex], [latex]b=-11[/latex] and [latex]c=3[/latex]. Hence
\begin{align*}
\Delta&=b^2-4ac\\
&=\left(-11 \right)^2-4\times 1\times 3\\
&=121-12\\
&=109.
\end{align*}
Solution b)
We have [latex]a=-3[/latex], [latex]b=0[/latex] and [latex]c=100[/latex]. Hence
\begin{align*}
\Delta&=b^2-4ac\\
&=0-4 \times \left(-3 \right)\times 100\\
&=1200.\\
\end{align*}
Solution c)
We have [latex]a=5[/latex], [latex]b=0[/latex] and [latex]c=0[/latex]. Hence
\begin{align*}
\Delta&=b^2-4ac\\
&=0-4\times \times 5 \times 0\\
&=0.\\
\end{align*}
2. Does the graph of [latex]y=27x^2-9x+250[/latex] cut the [latex]x[/latex]-axis? If so, at how many points?
Solution:
The discriminant
\begin{align*}
\Delta&=\left(-9 \right)^2-4\times 27 \times 250\\
&=81-27000\\
&\lt 0.\\
\end{align*}
As the discriminant is less than zero, the graph does not cut the [latex]x[/latex]-axis.
Try the following exercises.
Exercises: The Discriminant
1. Evaluate the discriminant of the following quadratic expressions:
\begin{align*}
a)\ x^2-2x+3\qquad b)\ -3x^2 \qquad c)\ 5x^2-3 \qquad \qquad\\
\end{align*}
Answers
[latex]\qquad a)\ -8\qquad b)\ 0 \qquad c)\ 60[/latex]
2. Does the graph of [latex]y=3x^2-5x-6[/latex] intersect the [latex]x[/latex]-axis and if so, how many times?
Answer
Yes. It cuts the [latex]x[/latex]-axis twice.
Quadratic equations
A general quadratic equation has the form
\begin{align*}
ax^2+bx+c &=0\\
\end{align*}
where [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are real constants. In a quadratic equation, the variable is raised to the power of two.
A quadratic equation will have no solutions, one real solution, or two real solutions. The number of solutions is determined by the value of the discriminant.
Quadratic equations may be solved by factorising the quadratic expression, completing the square, or using the quadratic formula. You should review the factorising quadratics section of the chapter on Factorisation before going further.
Solution by factorisation
Suppose you want to solve the quadratic equation
\begin{align*}
x^2-5x+6&=0.\\
\end{align*}
The left-hand side may be factorised as [latex]\left(x-3\right)\left(x-2\right)[/latex] and so
\begin{align*}
\left(x-3\right)\left(x-2\right)&=0.\\
\end{align*}
For this to be true, either [latex]\left(x-3\right) =0[/latex] or [latex]\left(x-2\right)=0[/latex]. Hence the solutions to the equation [latex]x^2-5x+6=0[/latex] are [latex]x=2[/latex] and [latex]x=3[/latex].
As a further example, consider the equation
\begin{align*}
6x^2+13x+5&=0.\\
\end{align*}
The left-hand side may be factorised as [latex]\left(2x+1\right)\left(3x+5\right)[/latex] and so
\begin{align*}
\left(2x+1\right)\left(3x+5\right)&=0.\\
\end{align*}
Hence the solutions to the equation are [latex]x=-\dfrac{1}{2}[/latex] and [latex]x=-\dfrac{5}{3}[/latex].
Examples: Solution by Factorisation
1. Solve [latex]x^2+4x+5=0[/latex].
Solution:
We can factorise to get:
\begin{align*}
x^2+4x+5&=\left(x-1 \right)\left(x+5\right).\\
\end{align*}
So if [latex]x^2+4x+5=0[/latex] we have [latex]x=1[/latex] or [latex]x=-5[/latex]. Hence, the solutions are [latex]x=1[/latex] and [latex]x=-5[/latex].
2. Factorise [latex]2x^2+x-1=0[/latex]
Solution:
Factorising we have
\begin{align*}
2x^2+x-1&=\left(2x-1\right)\left(x+1\right).\\
\end{align*}
Hence if [latex]2x^2+x+1=0[/latex], the solutions are [latex]x=\dfrac{1}{2}[/latex] and [latex]x=-1[/latex].
3. Solve [latex]6x^2-7x-3=0[/latex].
Solution:
The quadratic may be factorised as
\begin{align*}
6x^2-7x-3&=\left(2x-3\right)\left(3x+1\right).\\
\end{align*}
Hence, the solutions are [latex]x=\dfrac{3}{2}[/latex] and [latex]x=-\dfrac{1}{3}[/latex].
4. Solve the equation [latex]3x^2-5x=0[/latex].
Solution:
The quadratic function may be solved by taking out a factor of [latex]x[/latex] so that
\begin{align*}
0&=3x^2-5x\\
&=x\left(3x-5 \right).
\end{align*}
The solutions are: [latex]x=0[/latex] and [latex]x=\dfrac{5}{3}[/latex].
Exercises: Solution by factorisation
\begin{align*}
a)\ x^2+3x-10=0\qquad b)\ 3x^2+5x+2=0&\qquad c)\ 6x^2-11x+3=0\\
\end{align*}
Answers
a)\;x&=2,\ x=-5 \qquad b)\;x=-\dfrac{2}{3},\ x=-1 \qquad c)\;x=\dfrac{1}{3},\ x=\dfrac{3}{2}.\qquad \qquad \qquad \qquad \qquad \\
\end{align*}
Solution by completing the square
Sometimes it is not easy to factorise the quadratic equation by spotting linear factors. In such cases, the equation may be solved by completing the square on the quadratic function. Sometimes this involves the use of the difference of two squares formula:
\begin{align*}
a^2-b^2&=\left( a+b \right)\left( a-b \right).\\
\end{align*}
We give some examples below.
Examples: Solution by completing the square
1. Solve the equation [latex]x^2-5x+6 =0[/latex] by completing the square.
Solution:
We write
\begin{align*}
0&=x^2-5x+6 \\
&=x^2-5x+\left( \frac{5}{2} \right)^2-\left( \frac{5}{2} \right)^2+6\\
&=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}+6\\
&=\left(x-\frac{5}{2}\right)^2-\frac{1}{4}.\\
\end{align*}
Note that the last term can be written as [latex]\left(\dfrac{1}{2}\right)^2[/latex] and so
\begin{align*}
0&=\left(x-\frac{5}{2}\right)^2-\frac{1}{4}\\
&=\left(x-\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^{2}.
\end{align*}
Applying the difference of two squares gives:
\begin{align*}
0&=\left(x-\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2\\
&=\left(x-\frac{5}{2}+\frac{1}{2}\right)\left(x-\frac{5}{2}-\frac{1}{2}\right)\\
&=\left(x-\frac{4}{2}\right)\left(x-\frac{6}{2}\right)\\
&=\left(x-2 \right)\left(x-3 \right).\\
\end{align*}
Hence the solutions to the equation are [latex]x=2[/latex] and [latex]x=3[/latex].
2. Solve the equation [latex]3x^2-5x-4 =0[/latex] by completing the square.
Solution:
As the left-hand side is zero, we can divide through by [latex]3[/latex] to get:
\begin{align*}
0&=3x^2-5x-4\\
&=x^2-\frac{5}{3}x-\frac{4}{3}\\
&=x^2-\frac{5}{3}x+\left(\frac{5}{6}\right)^2-\left(\frac{5}{6}^2\right)-\frac{4}{3}\\
&=\left(x-\frac{5}{6} \right)^2-\frac{25}{36}-\frac{4}{3}\\
&=\left(x-\frac{5}{6} \right)^2-\frac{25}{36}-\frac{48}{36}\\
&=\left(x-\frac{5}{6} \right)^2-\frac{73}{36}\\
&=\left(x-\frac{5}{6}-\frac{\sqrt{73}}{6}\right)\left(x-\frac{5}{6}+\frac{\sqrt{73}}{6} \right)\\
&=\left(x-\frac{5-\sqrt{73}}{6}\right)\left(x-\frac{5+\sqrt{73}}{6} \right).\\
\end{align*}
Hence the solutions are [latex]x=\dfrac{5-\sqrt{73}}{6}[/latex] and [latex]x=\dfrac{5+\sqrt{73}}{6}[/latex] .
Here are some exercises to try.
Exercises: Solution by completing the square
1. Solve the following equations by completing the square.
\begin{align*}
&a)\ x^2-7x+2=0\qquad b)\ 3x^2+5x+2=0 \qquad c)\ 2x^2-3x-5=0
\end{align*}
Answers
\begin{align*}
1a)\ x&=\dfrac{7+ \sqrt{41}}{2}\quad x=\dfrac{7- \sqrt{41}}{2}\qquad \qquad \qquad \qquad\qquad \qquad\qquad \\
1b)\ x&=-1\quad x=-\dfrac{2}{3}\\
1c)\ x&=-1\quad x=\dfrac{5}{2}\\
\end{align*}
Solution using the quadratic formula
Sometimes it is not easy to factorise the quadratic equation. A solution may be found for any quadratic equation using the quadratic formula which is
\begin{align*}
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},\\
\end{align*}
where [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are the coefficients of the quadratic equation [latex]ax^2+bx+c=0[/latex]. As we saw in the previous section, this formula is derived by completing the square on the general quadratic.
For example, the previously solved equation
\begin{align*}
x^2-5x+6&=0,\\
\end{align*}
has [latex]a=1[/latex], [latex]b=-5[/latex] and [latex]c=6[/latex]. Hence the solution, using the quadratic formula, is
\begin{align*}
x=\frac{b\pm \sqrt{b^2-4ac}}{2a}\\
&=\frac{-\left(-5\right) \pm \sqrt{\left(-5\right)^2-4\times 1 \times6}}{2\times 1}\\
&=\frac{5\pm \sqrt{25-24}}{2}\\
&=\frac{5\pm 1}{2}\\
&=\frac{5+ 1}{2} \text{or} \frac{5- 1}{2}\\
&=3 \text{ or } 2.
\end{align*}
Hence, the solutions are [latex]x=2[/latex] and [latex]x=3[/latex] as before. Note that factorisation was the easier method of solution for this equation.
Examples: Solutions using the quadratic formula
1. Solve the equation [latex]x^2-3x-1=0[/latex].
Solution:
Here, [latex]a=1[/latex], [latex]b=-3[/latex] and [latex]c=-1[/latex]. Substituting in the quadratic formula:
\begin{align*}
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
&=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\times 1 \times \left(-1\right)}}{2\times 1}\\
&=\frac{3\pm \sqrt{9+4}}{2}\\
&=\frac{3\pm \sqrt{13}}{2}.
\end{align*}
Hence the solutions are [latex]\dfrac{3+ \sqrt{13}}{2}[/latex] and [latex]\dfrac{3- \sqrt{13}}{2}[/latex].
2. Solve [latex]3x^2+5x-2=0[/latex].
Solution:
Here, [latex]a=3[/latex], [latex]b=5[/latex] and [latex]c=-2[/latex]. Substituting in the quadratic formula:
\begin{align*}
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
&=\frac{-5\pm \sqrt{5^2-4\times 3 \times \left(-2\right)}}{2\times 3}\\
&=\frac{-5\pm \sqrt{25+24}}{6}\\
&=\frac{-5\pm \sqrt{49}}{6}\\
&=\frac{-5\pm7}{6}.
\end{align*}
Hence the solutions are [latex]\dfrac{-5+ 7}{6}=\dfrac{1}{3}[/latex] and [latex]\dfrac{-5- 7}{6}=-2.[/latex]
3. Solve [latex]x^2+2x+5=0[/latex].
Solution:
Here, [latex]a=1[/latex], [latex]b=2[/latex] and [latex]c=5[/latex]. Substituting in the quadratic formula:
\begin{align*}
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
&=\frac{-1\pm \sqrt{2^2-4\times 1 \times 5}}{2\times 1}\\
&=\frac{-1\pm \sqrt{4-20}}{2}\\
&=\frac{-1\pm \sqrt{-16}}{2}.\\
\end{align*}
Note that the square root term is negative. Since we cannot take the square root of a negative number, there are no solutions to this equation.
4. Solve [latex]24x^2-45x-50=0[/latex].
Solution:
Here, [latex]a=24[/latex], [latex]b=-45[/latex] and [latex]c=-50[/latex]. Substituting in the quadratic formula:
\begin{align*}
x&=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\
&=\frac{-\left( -45 \right) \pm \sqrt{ \left(-45 \right)^2-4\times 24 \times \left(-50 \right) }}{2 \times 24}\\
&=\frac{45 \pm \sqrt{ 6825}}{48}\\
&=\frac{45 \pm 5\sqrt{ 273}}{48}.\\
\end{align*}
Hence the solutions are [latex]x= \dfrac{45+ 5\sqrt{273}}{48}\approx 2.66[/latex] and [latex]x=\dfrac{45- 5\sqrt{273}}{48}\approx -0.78.[/latex]
Here are some exercises to try.
Exercises: Solutions using the quadratic formula
Solve the following equations using the quadratic formula.
\mathbf{1.}\ x^2+3x-10=0\qquad \mathbf{2.}\ 3x^2+5x+2=0&\qquad \mathbf{3.}\ 6x^2+7x-3=0\\
\end{align*}
Answers
\mathbf{1.}\ x&=2,\ x=-5 \qquad \mathbf{2.}\ x=-\dfrac{2}{3},\ x=-1 \qquad \mathbf{3.}\ x=\dfrac{1}{3},\ x=-\dfrac{3}{2} \qquad \\
\end{align*}
Graphical interpretation
We found earlier that the solutions to the equation [latex]x^2-5x+6=0[/latex] were [latex]x=2[/latex] and [latex]x=3[/latex]. Figure 8.5 shows the graph of [latex]y=x^2-5x+6[/latex].
Note that the solutions to the equation are the [latex]x[/latex]- intercepts of the graph. This is true in general. That is, the solutions to a quadratic equation are the [latex]x-[/latex] intercepts of the graph of [latex]y=ax^2+bx+c[/latex]. Finding the [latex]x-[/latex] intercepts is very useful for graphing quadratic functions.
Graphing a general quadratic function
It is always possible to plot the points on the graph of a quadratic function as we did earlier. However, it is easier to determine the turning point and the intercepts of the graph with the [latex]x[/latex] and [latex]y[/latex] axes. Knowing these points, and that the graph is symmetrical about the vertical line through the turning point, a graph of sufficient accuracy may be obtained.
The [latex]y[/latex]-intercept of the graph of [latex]y=ax^2+bx+c[/latex] is obtained by setting [latex]x=0[/latex].
The [latex]x[/latex]-intercepts, if there are any, can be found by solving [latex]ax^2 +bx+c=0[/latex] by factorising, completing the square or the quadratic formula. Unless the quadratic is of simple form, it is best to complete the square. Completing the square gives the turning point and the maximum or minimum value. If the graph cuts the x-axis twice, you will need to use the difference of two squares formula
\begin{align*}
a^2-b^2&=\left( a+b \right)\left( a-b \right)
\end{align*}
after completing the square to get the [latex]x[/latex]-intercepts.
Examples: Graphing quadratics
1. Sketch the graph of [latex]y=x^2-5x+6[/latex].
Solution 1: Simple factorisation.
First calculate the determinant.
\begin{align*}
\Delta&=b^2-4ac\\
&=\left(-5 \right)^2-4\times 1 \times 6\\
&=25-24\\
&=1.
\end{align*}
As [latex]\Delta>0[/latex] there will be two [latex]x[/latex]-intercepts. To find them we solve [latex]0=x^2-5x+6[/latex]. This factorises easily to give
\begin{align*}
0&=\left(x-3 \right)\left(x-2 \right).\\
\end{align*}
Hence the [latex]x[/latex]-intercepts are at [latex]x=2[/latex] and [latex]x=3[/latex].
The [latex]y[/latex]-intercept occurs when [latex]x=0[/latex], that is at [latex]y=0^2-5\times 0+6=6[/latex].
The graph will be concave up and so has a minimum. The minimum occurs at the turning point which occurs at [latex]x=-\dfrac{b}{2a}=2.5[/latex]. Note that we could also get the [latex]x[/latex]-coordinate of the turning point by noting it is midway between the two [latex]x[/latex]-intercepts: [latex]2[/latex] and [latex]3[/latex]. At the turning point, the function has the value
\begin{align*}
y&=2.5^2-5\left(2.5\right)+6\\
&=5.25-12.5+6\\
&=-0.25.
\end{align*}
Plotting these points gives Fig.8.6 below.
Solution 2: Completing the square.
We have\begin{align*}
x^2-5x+6&=x^2-5x+\left( \frac{5}{2} \right)^2 - \left( \frac{5}{2} \right)^2 +6\\
&=\left(x-\frac{5}{2} \right)^2 -\frac{25}{4}+\frac{24}{4}\\
&=\left(x-\frac{5}{2} \right)^2 -\frac{1}{4}.\\
\end{align*}From this, we see the turning point is at [latex]x=\dfrac{5}{2}=2.5[/latex], [latex]y=-\dfrac{1}{4}=-0.25[/latex] as before. To get the [latex]x[/latex]-intercepts we note that [latex]\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2[/latex] and use the difference of two squares rule:\begin{align*}
\left(x-\frac{5}{2} \right)^2 -\frac{1}{4}&=\left(x-\frac{5}{2} \right)^2-\left(\dfrac{1}{2}\right)^2\\
&=\left(x-\frac{5}{2}-\frac{1}{2} \right)\left(x-\frac{5}{2} +\frac{1}{2}\right)\\
&=\left(x-\frac{4}{2} \right) \left(x-\frac{6}{2} \right)\\
&=\left(x-2 \right) \left(x-3\right).\\
\end{align*}Setting this expression equal to zero, we find the [latex]x[/latex]-intercepts to be [latex]x=2[/latex] and [latex]x=3[/latex] as before.
Solution:
The determinant is\begin{align*}
\Delta&=b^2-4ac\\
&=11 ^2-4\times \frac{1}{2} \times 6\\
&=121-12\\
&=109.
\end{align*}As the determinant is greater than zero, we will have two [latex]x[/latex]-intercepts. Completing the square,\begin{align*}
y&=\frac{1}{2}x^2+11x+6\\
&=\frac{1}{2} \left[ x^2 +22x+12 \right] \\
&=\frac{1}{2} \left[ x^2 +22x+11^2-11^2+12 \right] \\
&=\frac{1}{2} \left[ \left( x+11 \right)^2 -121+12 \right] \\
&=\frac{1}{2} \left[ \left( x+11 \right)^2 -109 \right]. \\
\end{align*}We see that the turning point is at [latex]x=-11[/latex] and the value there is [latex]y=-109/2=-54.5[/latex]. As the coefficient of [latex]x^2[/latex] is greater than zero, the graph is concave up and the turning point is a minimum.The [latex]x[/latex]-intercepts may be found by setting [latex]y=0[/latex] and using the difference of two squares:\begin{align*}
0&=\frac{1}{2}\left[ \left(x+11 \right)^2 -109 \right]\\
&= \left(x+11-\sqrt{109} \right) \left(x+11+\sqrt{109} \right) \\
&\approx \left(x+0.5597 \right) \left(x+21.4403 \right). \\
\end{align*}Hence the [latex]x[/latex]-intercepts are [latex]x=-0.5597[/latex] and [latex]x=-21.4403[/latex].The y-intercept is found by setting [latex]x=0[/latex]:\begin{align*}
0&=\frac{1}{2}0^2+11\times 0+6\\
&=6.\\
\end{align*}The graph is shown in Fig. 8.7
3. Sketch the graph of [latex]y=-x^2+4x-5[/latex].
Solution:
The determinant is\begin{align*}
\Delta&=b^2-4ac\\
&=4^2-4\times \left(-1 \right) \times \left(-5 \right)\\
&=16-20\\
&=-4.
\end{align*}As the determinant is less than zero, there are no [latex]x[/latex]- intercepts. To find the turning point, complete the square:\begin{align*}
y&=-x^2+4x-5\\
&=-\left[ x^2 -4x + 5 \right]\\
&=-\left[ x^2 -4x +4-4+ 5 \right]\\
&=-\left[ \left(x-2 \right)^2 -4+ 5 \right] \\
&=-\left[ \left(x-2 \right)^2 +1 \right]. \\
\end{align*}The turning point occurs at [latex]x=2[/latex] and has value [latex]-1[/latex]. As the coefficient of [latex]x^2[/latex] is less than zero, the graph is concave down and the turning point is a maximum.Setting [latex]x=0[/latex] in the original expression we obtain the [latex]y[/latex]-intercept at\begin{align*}
y&=-0^2+4\times 0-5\\
&=-5.\\
\end{align*}
4. Sketch the graph of [latex]y=12x-9x^2-4[/latex].
Solution:
The determinant is
\begin{align*}
\Delta&=b^2-4ac\\
&=12^2-4\times \left(-9 \right) \times \left(-4\right)\\
&=144-144\\
&=0.
\end{align*}
y&=-9x^2+12x-4\\
&=-9\left[ x^2-\frac{12}{9}x +\frac{4}{9} \right]\\
&=-9\left[ x^2-\frac{12}{9}x +\left( \frac{6}{9} \right)^2 -\left( \frac{6}{9} \right)^2 +\frac{4}{9} \right]\\
&=-9\left[ \left(x-\frac{6}{9} \right)^2 - \frac{36}{81} +\frac{36}{81} \right]\\
&=-9\left(x-\frac{2}{3} \right)^2 . \\
\end{align*}
y&=-9x^2+12x-4\\
&=-9\times 0^2 + 12 \times 0 - 4\\
&=-4.
\end{align*}Hence the graph is as shown in Fig. 8.9 below.
Try these exercises.
Exercises: Graphing quadratics
1. Sketch the graph of [latex]y=x^2-7x+10[/latex]. Make sure you find and label all intercepts and the turning point.
Answer
2. Sketch the graph of [latex]y=4x^2+12x+9[/latex]. Make sure you find and label all intercepts and the turning point.
Answer
3. Sketch the graph of [latex]y=-6x^2+3x+5[/latex]. Make sure you find and label all intercepts and the turning point.
Answer
Higher order equations
The techniques for solving quadratic equations may be used with higher-order equations that can be arranged in a quadratic form. For example, the fourth-order equation
\begin{align*}
x^4-5x^2+6&=0
\end{align*}
is quadratic in [latex]x^2[/latex]. To solve it, set [latex]m=x^2[/latex] then
\begin{align*}
m^2-5m+6&=0\\
\implies \left(m-2\right)\left(m-3\right)&=0\\
\end{align*}
which has solutions [latex]m=2[/latex] and [latex]m=3[/latex]. Hence the solution to the original equation is [latex]x=\pm\sqrt{2}[/latex] and [latex]x=\pm \sqrt{3}[/latex]. The solutions are the [latex]x[/latex]- intercepts of the graph of [latex]y=x^4-5x^2+6[/latex] as shown in Fig. 8.10.
Examples: Higher order equations
1. Solve the equation [latex]x^4-5x^2+4=0[/latex].
Solution:
Let [latex]m=x^2[/latex], then the equation may be written as
\begin{align*}
m^2-5m+4&=0.\\
\end{align*}
Factorising we have
\begin{align*}
m^2-5m+4&=\left(m-1\right)\left(m-4\right).
\end{align*}
Hence
\begin{align*}
m^2-5m+4&=0\\
\implies \left(m-1\right)\left(m-4\right)=0.
\end{align*}
So either [latex]m=1[/latex] or [latex]m=4[/latex].
If [latex]m=1[/latex] then [latex]x^2=1[/latex] and [latex]x=\pm 1[/latex]. If [latex]m=4[/latex] then [latex]x=\pm 2[/latex].
Hence the solutions are [latex]x=-2[/latex], [latex]x=-1[/latex], [latex]x=1[/latex] and [latex]x=2[/latex].
As in the case of quadratic equations, these solutions are the [latex]x[/latex]- intercepts of the graph of [latex]y=x^4-5x^2+4[/latex] as shown in Fig. 8.11.
2. Solve the equation [latex]5a^4-2a^2-7=0[/latex].
Solution:
Let [latex]x=a^2[/latex] then the equation may be written as
\begin{align*}
5x^2-2x-7&=0.\\
\end{align*}
Factorising we have
\begin{align*}
5x^2-2x-7&=\left(5x-7\right)\left(x+1\right)
\end{align*}
and so either [latex]x=\dfrac{7}{5}[/latex] or [latex]x=-1[/latex].
If [latex]x=7/5[/latex], then
\begin{align*}
a^2&=\frac{7}{5}\\
\implies a&=\pm \sqrt{\frac{7}{5}}.\\
\end{align*}
If [latex]x=-1[/latex] we have [latex]a^2=-1[/latex] which is not possible. Hence the solutions are [latex]x=\sqrt{7/5}[/latex] and [latex]x=-\sqrt{7/5}[/latex].
In this case, the graph only cuts the [latex]x[/latex]- axis at two points as shown in Fig. 8.12.
Here are some exercises to try.
Exercises: Higher order equations
\mathbf{1.}\ x^4+3x^2-10=0\qquad \mathbf{2.}\ 3x^4+5x^2+2=0&\qquad \mathbf{3.}\ 6x^4-11x^2+3=0\\
\end{align*}
Answers
&\mathbf{1.}\ x=-\sqrt{2}, x=\sqrt{2}\\
&\mathbf{2.}\ \text{No solutions} \\
&\mathbf{3.}\ x=-\dfrac{1}{\sqrt{3}},\ x=\dfrac{1}{\sqrt{3}},\ x=-\sqrt{\dfrac{3}{2}},\ x=\sqrt{\dfrac{3}{2}}. \qquad \qquad \qquad \qquad \quad \quad
\end{align*}
Exponential quadratic equations
Consider the equation
\begin{align*}
e^{2x}-5e^x+6&=0.\\
\end{align*}
This is also of quadratic form. Setting [latex]m=e^x[/latex] it may be written:
\begin{align*}
m^{2}-5m+6&=0\\
\implies \left(m-2\right)\left(m-3\right)&=0\\
\end{align*}
which has solutions [latex]m=2[/latex] and [latex]m=3[/latex]. For [latex]m=2[/latex] we have
\begin{align*}
e^x&=2.\\
\end{align*}
Taking natural logs of both sides gives
\begin{align*}
x&=\ln\left(2\right).\\
\end{align*}
For [latex]m=3[/latex] we have
\begin{align*}
e^x&=3\\
\implies x&=\ln\left(3\right).\\
\end{align*}
Hence the solutions to
\begin{align*}
e^{2x}-5e^x+6&=0\\
\end{align*}
are [latex]x=\ln\left(2\right)[/latex] and [latex]x=\ln\left(3\right)[/latex]. The solutions are the [latex]x-[/latex] intercepts of the graph of [latex]y=e^{2x}-5e^x+6[/latex].
To find the [latex]y[/latex]-intercept, set [latex]x=0[/latex] to get
\begin{align*}
y&=e^{2\times 0}-5e^0+6\\
&=1-5+6\\
&=2.
\end{align*}
These points are shown in Fig 8.13.
.
Examples: Exponential quadratic equations
1. Solve the equation [latex]3e^{2x}-7e^x+2=0[/latex].
Solution:
Let [latex]m=e^x[/latex], then the equation becomes
\begin{align*}
3m^{2}-7m+2&=0\\
\implies \left(3m-1\right)\left(m-2\right)&=0.\\
\end{align*}
Hence [latex]m=1/3[/latex] or [latex]m=2[/latex].
If [latex]m=1/3[/latex],
\begin{align*}
e^x&=\frac{1}{3}\\
\implies x&=\ln\left(\frac{1}{3}\right).\\
\end{align*}
If [latex]m=2[/latex],
\begin{align*}
e^x&=2\\
\implies x&=\ln \left(2 \right).\\
\end{align*}
Hence the solutions are [latex]x=\ln\left(1/3 \right)[/latex] and [latex]x=\ln \left( 2 \right)[/latex].
2. Find [latex]x[/latex] such that [latex]5e^{2x}-11e^x-21=0[/latex].
Solution:
Let [latex]m=e^x[/latex], then the equation becomes
\begin{align*}
5m^{2}-11m-21&=0.\\
\end{align*}
There is no simple factorisation of this equation. We must use completing the square or the quadratic formula. In this example, we use the quadratic formula with [latex]a=5[/latex], [latex]b=-11[/latex] and [latex]c=-21[/latex]. Hence
\begin{align*}
m&=\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\times5 \times \left(-21 \right)}}{2\times 5}\\
&=\frac{11 \pm \sqrt{121+420}}{10}\\
&=\frac{11 \pm\sqrt{541}}{10}.\\
\end{align*}
Hence [latex]m=\dfrac{ 11-\sqrt{541}}{10}[/latex] or [latex]m=\dfrac{ 11+\sqrt{541}}{10}[/latex].
The solution [latex]m=\dfrac{ 11-\sqrt{541}}{10}[/latex] is less than zero and implies [latex]e^x \lt 0[/latex]. As the exponential function only takes positive values, we ignore this solution.
If [latex]m=\dfrac{ 11+\sqrt{541}}{10}[/latex], then
\begin{align*}
e^x&=\frac{ 11+\sqrt{541}}{10}\\
\implies x&=\ln\left(\frac{ 11+\sqrt{541}}{10}\right).\\
\end{align*}
Hence the solution is [latex]x=\ln\left(\dfrac{ 11+\sqrt{541}}{10}\right)[/latex].
Here are some exercises to try.
Exercises: Exponential quadratic equations
Solve the equations.
\begin{align*}
\mathbf{1.}\ e^{2x}+3e^x-10=0\qquad \mathbf{2.}\ 6e^{2x}-11e^x+3=0 \qquad \qquad \qquad \qquad \qquad\\
\end{align*}
Answers:
[latex]\begin{array}{lllllllllllllll} \quad \mathbf{1.}\;x=\ln\left( 2 \right)& & \mathbf{2.}\;\ln\left( 1/3 \right)=-\ln\left(3 \right),\ x=\ln\left( 3/2 \right). \end{array}[/latex]
Key takeaways
- A quadratic equation has the form
\begin{align*}
ax^2+bx+c&=0 \quad a,\,b\, \text{and } c \text{ are constants.}\\
\end{align*}
- The turning point of a quadratic occurs at [latex]x=-\dfrac{b}{2a}.[/latex]
- Quadratic equations may be solved using factorisation (if possible), completing the square or the quadratic formula.
- The quadratic formula is
\begin{align*}
x&=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\\
\end{align*}
and gives the solution of the general quadratic equation [latex]ax^2+bx+c=0.[/latex]
- The quantity [latex]\Delta =b^2 -4ac[/latex] is called the discriminant.
- If [latex]\Delta \lt 0[/latex], there are no real solutions to the quadratic equation. This means the graph of [latex]y=ax^2+bx+c[/latex] does not touch or cut the x- axis.
- If [latex]\Delta=0[/latex], there is one real solution to the quadratic equation. This means the graph of [latex]y=ax^2+bx+c[/latex] touches the [latex]x-[/latex] axis at one point.
- If [latex]\Delta>0[/latex], there are two real solutions to the quadratic equation. This means the graph of [latex]y=ax^2+bx+c[/latex] cuts the [latex]x-[/latex] axis at two points.
- Some higher-order equations or exponential equations may be reduced to a quadratic equation.
