Learning Objectives
- Determine the zeros of a given quadratic function by factoring.
- Determine the zeros of a given quadratic function by completing the square
- Determine the zeros of a given quadratic function by using the quadratic formula
- Understand the relationship between zeros and [latex]x[/latex]-intercepts
- Understand that quadratic functions with complex zeros have no [latex]x[/latex]-intercepts
Factoring
In chapter 3, we learned several factoring methods and used them to find the zeros of polynomial functions. A quadratic function is a special case of a polynomial function. It is a trinomial of the form [latex]f(x)=ax^2+bx+c[/latex], where [latex]a, b, c[/latex] are real numbers and [latex]a\ne 0[/latex]. Consequently, we have already learned how to determine the zeros of a quadratic function by factoring.
For example, to determine the zeros of [latex]f(x)=9x^2-4[/latex], we set [latex]f(x)=0[/latex] and solve for [latex]x[/latex] by factoring and using the zero-product property:
[latex]\begin{aligned}9x^2-4 &=0\\(3x-2)(3x+2)&=0\\3x-2=0\text{ or }3x+2&=0\\x=\dfrac{2}{3}\text{ or }x&=-\dfrac{2}{3}\end{aligned}[/latex]
The zeros of the function [latex]f(x)=9x^2-4[/latex] are [latex]x=-\dfrac{2}{3},\;\dfrac{2}{3}[/latex] or [latex]x=\pm\dfrac{2}{3}[/latex], where the symbol [latex]\pm[/latex] is read “plus or minus” or “positive or negative”.
We also learned that the zeros correspond with the [latex]x[/latex]-coordinates of the [latex]x[/latex]-intercepts on the graphs of the functions. So the [latex]x[/latex]-intercepts of [latex]f(x)=9x^2-4[/latex] are [latex]\left(-\dfrac{2}{3},\;0\right)[/latex] and [latex]\left(\dfrac{2}{3}, 0\right)[/latex].
We will not repeat those factoring methods here. Factoring and using the zero-product property works well when the function factors. But in most cases, functions do not factor, so other methods need to be employed. In this section, we will focus on two methods that can be used to determine the zeros (and consequently the [latex]x[/latex]-intercepts) of any quadratic function that may or may not factor.
Completing the Square
We just learned in section 4.3 how to convert a quadratic function from its standard form [latex]f(x)=ax^2+bx+c[/latex] into its vertex form [latex]f(x)=a(x-h)^2+k[/latex] to determine the vertex of the function. In this section, we will take advantage of that new found knowledge to determine the zeros of a quadratic function.
For example, to determine the zeros of the function [latex]f(x)=x^2 - 4[/latex] we set the function equal to 0. To solve the equation, we may isolate the perfect square, so we have a perfect square equal to a constant:
[latex]\begin{aligned}x^2 - 4 &= 0\\x^2&=4\end{aligned}[/latex]
To solve this equation we first need to learn about the square root property.
The Square Root Property
The square root property of equality tells us that if we have an equation [latex]A=B[/latex], we can take the square root of each side of the equation and get an equivalent equation, [latex]\sqrt{A}=\sqrt{B}[/latex].
The Square Root Property of Equality
If [latex]A=B[/latex], then [latex]\sqrt{A}=\sqrt{B}[/latex]
So, since we have [latex]x^2=4[/latex], then [latex]\sqrt{x^2}=\sqrt{4}[/latex].
We already know that [latex]\sqrt{4}=2[/latex], but what is [latex]\sqrt{x^2}[/latex]?
There are two cases to consider: [latex]x≥0[/latex] and [latex]x<0[/latex].
Suppose [latex]x=9[/latex], then [latex]x^2=(9)^2=81[/latex]. Then [latex]\sqrt{x^2}=\sqrt{81}=9[/latex]. And since [latex]9=x[/latex], [latex]\sqrt{x^2}=x[/latex] when [latex]x=9[/latex].
In fact, we could use any value of [latex]x≥0[/latex] and get the same result. So, [latex]\sqrt{x^2}=x[/latex] when [latex]x≥0[/latex].
But what happens when [latex]x<0[/latex]?
Suppose [latex]x=-7[/latex], then [latex]x^2=(-7)^2=49[/latex]. Then [latex]\sqrt{x^2}=\sqrt{49}=7[/latex]. But [latex]x=-7[/latex], not [latex]7[/latex]! So when [latex]x<0[/latex], [latex]\sqrt{x^2}=-x[/latex], which is a positive answer when [latex]x[/latex] is negative.
Putting theses two results together, and not knowing whether [latex]x[/latex] is positive or negative, we have, [latex]\sqrt{x^2}=| x |[/latex].
SQUARE ROOTS
[latex]\sqrt{x^2}=| x |[/latex]
Going back to our example,
[latex]\begin{aligned}x^2&=4\\\sqrt{x^2}&=\sqrt{4}\\|x|&=2\\x&=\pm2\end{aligned}[/latex]
Therefore, the zeros are [latex]x=-2[/latex] and [latex]x=2[/latex]. These correspond with the two [latex]x[/latex]-intercepts on the graph of the function (2, 0) and (–2, 0).
Notice that we have an absolute value equal to a constant, (i.e., [latex]|x|=2[/latex]). This means that [latex]x[/latex] will equal either positive or negative of the constant, (i.e., [latex]x=\pm2[/latex]). This is because the absolute value of a constant and its opposite are the same, (i.e., [latex]|2|=|-2|=2[/latex].
Most often, we will not have a perfect square equal to a perfect square. For example, we may end up solving [latex]x^2=72[/latex], where 72 is not a perfect square. Let’s recall how to deal with simplifying a square root that is not a perfect square.
Simplifying a Square Root
A square root can be simplified by separating perfect squares from non-perfect squares inside a square root. A number may be factored such that one or several perfect squares are among the factors. We can then use the product property of radicals to take the square root of the factors.
Product property of radicals
[latex]\sqrt{ab}=\sqrt{a}\sqrt{b}[/latex] for all [latex]a,\;b≥0[/latex]
For example, the number 72 may be factored as [latex]72=2^3\times 3^2[/latex]. There are two perfect squares, [latex]2^2[/latex] and [latex]3^2[/latex], in the prime factorization of 72. According to the product property of radicals, we may express [latex]\sqrt{72}[/latex] as the product of square roots:
[latex]\begin{aligned}\sqrt{72}&=\sqrt{2^3\times3^2}\\&=\sqrt{2\times2^2\times3^2}\\&=\sqrt{2}\times\sqrt{2^2}\times\sqrt{3^2}\,\,\,\,\,\,\,\text{Product property of radicals}\\&=\sqrt{2}\times2\times3\\&=6\sqrt{2}\end{aligned}[/latex]
Therefore, [latex]\sqrt{72}[/latex] may be simplified to [latex]6\sqrt{2}[/latex].
So, if we are solving the equation [latex]x^2=72[/latex], we can take the square root of both sides and simplify the radical:
[latex]\begin{aligned}x^2&=72\\\sqrt{x^2}&=\sqrt{72}\\|x|&=6\sqrt{2}\\x&=\pm6\sqrt{2}\end{aligned}[/latex]
Example 1
Simplify [latex]\sqrt{1080}[/latex].
Solution
[latex]\begin{aligned}\sqrt{1080}&=\sqrt{2^3\times3^3\times5}\\&=\sqrt{2\times2^2\times3\times3^2\times5}\\&=\sqrt{2^2\times3^2\times2\times3\times5}\\&=\sqrt{2^2}\times\sqrt{3^2}\times\sqrt{2\times3\times5}\,\,\,\,\,\,\,\text{Product property of radicals}\\&=2\times3\times\sqrt{30}\\&=6\sqrt{30}\end{aligned}[/latex]
Try It 1
Simplify [latex]\sqrt{220}[/latex].
Example 2
Solve: [latex]\left(x-5\right)^2=40[/latex]
Solution
[latex]\begin{aligned}\left(x-5\right)^2&=40\\\sqrt{\left(x-5\right)^2}&=\sqrt{40}&&\text{Square root both sides}\\|x-5|&=\sqrt{4\cdot10}&&\text{look for perfect square factors under the radical}\\|x-5|&=\sqrt{4}\sqrt{10}&&\text{Multiplication property of radicals}\\|x-5|&=2\sqrt{10}&&\text{Simplify the radicals}\\x-5&=\pm2\sqrt{10}&&\text{Remove the absolute value and add a }\pm\\x&=5\pm2\sqrt{10}&&\text{Add 5 to both sides}\end{aligned}[/latex]
Try It 2
Solve: [latex]\left(x+1\right)^2=32[/latex]
Let’s use our new found knowledge to determine zeros of quadratic functions.
Example 3
Determine the zeros of [latex]g(x)=4(x-3)^2 - 32[/latex], then state the [latex]x[/latex]-intercepts.
Solution
We start by setting [latex]g(x)=0[/latex], then isolating the perfect square:
[latex]\begin{aligned}4(x-3)^2 - 32 &= 0\\4(x-3)^2&=32\\(x-3)^2&=8\end{aligned}[/latex]
We can now take the square root of both sides of the equation:
[latex]\begin{aligned}\sqrt{(x-3)^2}&=\sqrt{8}\\|x-3|&=\sqrt{4\cdot 2}\\x-3&=\pm2\sqrt{2}\\x&=3\pm2\sqrt{2}\end{aligned}[/latex]
The zeros of the function are [latex]x=3+2\sqrt{2}[/latex] and [latex]x=3-2\sqrt{2}[/latex].
The [latex]x[/latex]-coordinates are therefore [latex]\left(3+2\sqrt{2},\,0\right)[/latex] and [latex]\left(3-2\sqrt{2},\,0\right)[/latex].
Example 4
Determine the zeros of [latex]f(x)=2(x+4)^2 - 50[/latex], then state the [latex]x[/latex]-intercepts.
Solution
Start by setting [latex]f(x)=0[/latex], the isolating the perfect square:
[latex]\begin{aligned}2(x+4)^2 - 50&=0\\2(x+4)^2&=50\\(x+4)^2&=25\end{aligned}[/latex]
We can now take the square root of both sides of the equation:
[latex]\begin{aligned}\sqrt{(x+4)^2}&=\sqrt{25}\\|x+4|&=5\\x+4&=\pm5\\x&=-4\pm5\end{aligned}[/latex]
This results in two equations:
[latex]\begin{aligned}x&=-4+5\\&=1\end{aligned}[/latex] and [latex]\begin{aligned}x&=-4-5\\&=-9\end{aligned}[/latex]
The zeros of the function are [latex]x=1[/latex] and [latex]x=-9[/latex].
The [latex]x[/latex]-intercepts are therefore [latex](1, 0)[/latex] and [latex](-9, 0)[/latex].
Try It 3
Determine the zeros of [latex]f(x)=3(x-1)^2 - 27[/latex], then state the [latex]x[/latex]-intercepts.
Try It 4
Determine the zeros of [latex]f(x)=2(x+3)^2 - 36[/latex], then state the [latex]x[/latex]-intercepts.
If the function is given in standard form, we first need to convert it to vertex form using completing the square.
Example 5
Determine the zeros of [latex]f(x)=2x^2+12x-20[/latex], then state the [latex]x[/latex]-intercepts.
Solution
Start by using completing the square to convert the function into vertex form:
[latex]\begin{aligned}f(x)&=2x^2+12x-20\\&=2\left(x^2+6x\right)-20\\&=2\left[(x+3)^2-9\right]-20\\&=2(x+3)^2-18-20\\&=2(x+3)^2-38\end{aligned}[/latex]
Now we can set the function equal to zero:
[latex]\begin{aligned}2(x+3)^2-38&=0\\2(x+3)^2&=38\\(x+3)^2&=19\\\sqrt{(x+3)^2}&=\sqrt{19}\\|x+3|&=\sqrt{19}\\x+3&=\pm\sqrt{19}\\x&=-3\pm\sqrt{19}\end{aligned}[/latex]
The zeros of the function are [latex]x=-3-\sqrt{19}[/latex] and [latex]x=-3+\sqrt{19}[/latex].
The [latex]x[/latex]-intercepts are therefore [latex](-3-\sqrt{19}, 0)[/latex] and [latex](-3+\sqrt{19}, 0)[/latex].
Try It 5
Determine the zeros of [latex]f(x)=x^2+4x-12[/latex], then state the [latex]x[/latex]-intercepts.
Try It 6
Determine the zeros of [latex]f(x)=3x^2+17x+10[/latex], then state the [latex]x[/latex]-intercepts.
Imaginary Numbers and Complex Zeros
While trying to determine the zeros of a quadratic function, it is possible that we end up with an equation where the perfect square is equal to a negative number. For example, determining the zeros of the the function [latex]f(x)=x^2+4[/latex] leads to the equation [latex]x^2 = -4.[/latex] Does this equation have a solution?
In the domain of real numbers, it is impossible to find a number whose square is a negative number. The square of every real number, whether positive or negative, is positive. Therefore, there are no real number solutions for this equation.
However, the square root of a negative number is defined in the set of imaginary numbers. The square root of –1 is defined as the letter [latex]i[/latex].
[latex]\sqrt{-1} = i[/latex]
Consequently, all negative real numbers have a square root that is an imaginary number:
[latex]\sqrt{-3} = \sqrt{-1}\cdot\sqrt{3} = i\sqrt{3}[/latex] (we always write the [latex]i[/latex] in front of the radical)
[latex]\sqrt{-100} = \sqrt{-1}\cdot\sqrt{100} = i\sqrt{100} = 10i[/latex]
When we combine through addition a real number [latex]a[/latex] and an imaginary number [latex]bi[/latex], where [latex]a,\;b[/latex] are real numbers, we form a complex number [latex]a+bi[/latex].
COMPLEX NUMBERS
The set of complex numbers [latex]\mathbb{C}[/latex] is the union of the set of real numbers [latex]\mathbb{R}[/latex] and the set of imaginary numbers [latex]\{bi\;|\;i=\sqrt{-1}\;,\;b\in \mathbb{R}\;\}[/latex].
For any real numbers [latex]a[/latex] and [latex]b[/latex], [latex]a+bi[/latex] is a complex number.
Complex zeros will show up when we end up with a perfect square equal to a negative number.
Example 6
Determine the zeros of the function [latex]f(x)=\left(x+3\right)^2+16[/latex].
Solution
To determine the zeros we set the function equal to zero and solve the equation:
[latex]\begin{aligned}(x+3)^2+16&=0\\(x+3)^2&=-16\\\sqrt{(x+3)^2}&=\sqrt{-16}\\|x+3|&=4i\\x+3&=\pm4i\\x&=-3\pm4i\end{aligned}[/latex]
The result is two complex zeros [latex]x=-3-4i[/latex] and [latex]x=-3+4i[/latex]
Example 7
Determine the zeros of the function [latex]g(x)=x^2-4x+24[/latex].
Solution
We first convert the function to vertex form by completing the square:
[latex]\begin{aligned}x^2-4x+24&=(x-2)^2-4+24\\&=(x-2)^2+20\end{aligned}[/latex]
To determine the zeros we set the function equal to zero and solve the equation:
[latex]\begin{aligned}(x-2)^2+20&=0\\(x-2)^2&=-20\\\sqrt{(x-2)^2}&=\sqrt{-20}\\|x-2|&=\sqrt{-4}\sqrt{5}\\x-2&=\pm2i\sqrt{5}\\x&=2\pm2i\sqrt{5}\end{aligned}[/latex]
The result is two complex zeros [latex]x=2-2i\sqrt{5}[/latex] and [latex]x=2+2i\sqrt{5}[/latex].
Try It 7
Determine the zeros of the function [latex]g(x)=x^2-6x+30[/latex].
A function having complex zeros is not unusual, but what does this mean for the [latex]x[/latex]-intercepts of the graph of such a function?
Let’s look at the graphs of the functions from examples 4 and 5.
[latex]g(x)=x^2+6x+25[/latex] with complex zeros [latex]x=-3-4i[/latex] and [latex]x=-3+4i[/latex] |
complex zeros [latex]x=2-2i\sqrt{5}[/latex] and [latex]x=2+2i\sqrt{5}[/latex] |
| Figure 1. Graphs of functions with complex zeros. | |
[latex]g(x)=x^2-4x+24[/latex] withThere are no [latex]x[/latex]-intercepts when a quadratic function has complex zeros. This is because the coordinate system, and in particular the [latex]x[/latex]-axis, graphs only real numbers.
graphs of quadratic functions with complex zeros
The graph of a quadratic function with complex zeros has no [latex]x[/latex]-intercepts; the parabola never intersects the [latex]x[/latex]-axis.
The Quadratic Formula
Another way to determine the zeros of a quadratic function [latex]f(x) = ax^2 + bx + c[/latex] is to use the quadratic formula.
The quadratic formula
For any quadratic function [latex]f(x)=ax^2+bx+c[/latex], the zeros of the function are given by
[latex]x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/latex]
To use the quadratic equation, we need a quadratic function in the general form [latex]f(x)=ax^2+bx+c[/latex] so that we can pick out the values of [latex]a,\;b,\;c[/latex].
For example if we are asked to determine the zeros of the function [latex]h(x)=3x^2-5x+1[/latex], we evaluate the quadratic equation when [latex]a = 3, b = -5, c = 1[/latex]:
[latex]\begin{aligned}x &=\dfrac{-(-5) \pm\sqrt{(-5)^2-4(3)(1)}}{2(3)} \\&= \dfrac{5\pm\sqrt{25-12}}{6} \\=& \dfrac{5\pm\sqrt{13}}{6} \end{aligned}[/latex]
Therefore, the zeros are [latex]x=\dfrac{5 - \sqrt{13}}{6}[/latex] and [latex]x=\dfrac{5 + \sqrt{13}}{6}[/latex].
If we were asked to find the [latex]x[/latex]-intercepts, they are [latex](\dfrac{5 + \sqrt{13}}{6}, 0)[/latex] and [latex](\dfrac{5 - \sqrt{13}}{6}, 0)[/latex].
Example 8
Determine the zeros of the function [latex]g(x)=x^2-4x+2[/latex]. Then state the [latex]x[/latex]-intercepts.
Solution
To use the quadratic formula we need [latex]a=1,\;b=-4,\;c=2[/latex]:
[latex]\begin{aligned}x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}\\&=\dfrac{4\pm\sqrt{16-8}}{2}\\&=\dfrac{4\pm\sqrt{8}}{2}\\&=\dfrac{4\pm2\sqrt{2}}{2}\\&=2\pm\sqrt{2}\\&=2+\sqrt{2},\;\;2-\sqrt{2}\end{aligned}[/latex]
The zeros of the function are [latex]x=2+\sqrt{2}[/latex] and [latex]x=2-\sqrt{2}[/latex].
The [latex]x[/latex]-intercepts are [latex](2+\sqrt{2}, 0)[/latex] and [latex](2-\sqrt{2}, 0)[/latex].
Example 9
Determine the zeros of the function [latex]g(x)=2x^2-x+5[/latex]. Then state the [latex]x[/latex]-intercepts.
Solution
First pick out [latex]a=2,\;b=-1,\;c=5[/latex], then evaluate the quadratic formula:
[latex]\begin{aligned}x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(2)(5)}}{2(2)}\\&=\dfrac{1\pm\sqrt{1-40}}{4}\\&=\dfrac{1\pm\sqrt{-39}}{4}\\&=\dfrac{1\pm i\sqrt{39}}{4}\\&=\dfrac{1}{4}\pm\dfrac{i\sqrt{39}}{4}\end{aligned}[/latex]
The zeros of the function are [latex]x=\dfrac{1}{4}-\dfrac{i\sqrt{39}}{4}[/latex] and [latex]\dfrac{1}{4}+\dfrac{i\sqrt{39}}{4}[/latex].
There are no [latex]x[/latex]-intercepts since the zeros are complex.
Try It 8
Determine the zeros of the function [latex]g(x)=x^2-3x+2[/latex]. Then state the [latex]x[/latex]-intercepts.
Try It 9
Determine the zeros of the function [latex]g(x)=3x^2+12[/latex]. Then state the [latex]x[/latex]-intercepts.
Proof of the Quadratic Formula
This quadratic formula may be derived by converting the function [latex]f(x) = ax^2 + bx + c[/latex] into vertex form by completing the square, and then setting [latex]f(x)=0[/latex] and solving for [latex]x[/latex]. The following proof shows how this formula is derived.
[latex]\begin{aligned}f(x)&=ax^2+bx+c\\&=a\left[x^2 + \frac{b}{a}x\right] + c\\&=a\left[\left(x+\left(\dfrac{b}{2a}\right)\right)^2 - \left(\dfrac{b}{2a}\right)^2\right] + c\\&=a\left(x+\left(\frac{b}{2a}\right)\right)^2 - a\left(\frac{b}{2a}\right)^2 + c\\&=a\left(x+\left(\frac{b}{2a}\right)\right)^2 - \frac{b^2}{4a} + c\end{aligned}[/latex]
To find the zeros, set the function equal to zero, and then solve for [latex]x[/latex]:
[latex]\begin{aligned}a\left(x+\left(d\frac{b}{2a}\right)\right)^2 - \dfrac{b^2}{4a} + c &= 0\\a\left(x+\left(\dfrac{b}{2a}\right)\right)^2&=\dfrac{b^2}{4a} - c\\a\left(x+\left(\dfrac{b}{2a}\right)\right)^2 &=\dfrac{b^2-4ac}{4a}\\x+\left(\dfrac{b}{2a}\right)^2 &= \dfrac{b^2-4ac}{4a^2}\\\sqrt{\left(x+\left(\dfrac{b}{2a}\right)\right)^2} &=\sqrt{\dfrac{b^2-4ac}{4a^2}}\\\mathopen| x+\dfrac{b}{2a}\mathclose| &=\dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}\\x+\dfrac{b}{2a}&=\pm\dfrac{\sqrt{b^2-4ac}}{2|a|}\\x &= -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{\pm2a}\\x &= -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x &= \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}\end{aligned}[/latex]
