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 to get an equivalent equation, [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].

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].

Notice that when we simplified [latex]\sqrt{8}[/latex] in Example 1, we rewrote [latex]\sqrt{8}[/latex] as [latex]\sqrt{4\cdot 2}[/latex]. We did this because to simplify square roots we look for perfect square factors in the radicand. Since [latex]4[/latex] is the largest perfect square factor of [latex]8[/latex], we write [latex]8=4\cdot 2[/latex]. We can then use the product property of radicals [latex]\left (\sqrt{ab}=\sqrt{a}\sqrt{b}\right )[/latex] to simplify: [latex]\sqrt{8}=\sqrt{4\cdot 2}=\sqrt{4}\sqrt{2}=2\sqrt{2}[/latex].

If the function is given in standard form, we first need to convert it to vertex form using completing the square.

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 zeros will show up when we end up with a perfect square equal to a negative number.

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]	[latex]g(x)=x^2-4x+24[/latex] with 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.

There 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.

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.

To use the quadratic equation, we need a quadratic function in the standard 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].

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]

Q.E.D.