ORCCA Strategies for Solving Quadratic Equations

Section 9.5 Strategies for Solving Quadratic Equations

In this section, we will review how to solve quadratic equations using three different methods: The square root method, factoring and the quadratic formula.

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Subsection 9.5.1 How to Choose a Method for Solving a Quadratic Equation

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Process 9.5.1.

So far, we have learned three methods for solving quadratic equations in standard form, $ax^2+bx+c=0\text{:}$

When $b=0\text{,}$ as in $x^2-4=0\text{,}$ we can use the square root property, Property 9.2.1.
If the quadratic equation can be written as a squared expression equal to a constant, as in $\left(x-3\right)^2=25\text{,}$ we can also use the square root property, Property 9.2.1.
If we can easily factor the polynomial, as in $x^2-4x-12=0\text{,}$ we will solve the equation by factoring and use the zero product property, Property 7.7.3.
If we cannot solve the equation by the first two methods, we must use the Quadratic Formula, Property 9.3.3. This formula works for any quadratic equation, but the first two methods are usually easier.

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Let's look at a few examples for how to choose which method to use.

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Example 9.5.2.

Solve for $y$ in $y^2-49=0\text{.}$

Explanation

In this equation, $b=0\text{,}$ so it is easiest to use the square root method. We isolate the squared quantity and then use the square root property.

\begin{align*} y^2-49\amp=0\\ y^2\amp=49 \end{align*}

\begin{align*} y\amp=-\sqrt{49} \amp\text{ or }\amp\amp y\amp=\sqrt{49}\\ y\amp=-7\amp\text{ or }\amp\amp y\amp=7 \end{align*}

The solution set is $\{-7,7\}\text{.}$

Because $49$ is a perfect square, we could also solve this equation by factoring.

\begin{align*} y^2-49\amp=0\\ (y+7)(y-7)\amp=0 \end{align*}

\begin{align*} y+7\amp=0\amp\text{ or }\amp\amp y-7\amp=0\\ y\amp=-7\amp\text{ or }\amp\amp y\amp=7 \end{align*}

We get the same solution set, $\{-7,7\}\text{.}$

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We can also use the square root method when a binomial is squared, like

$(p-1)^2\text{,}$ as we will see in the next example.

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Example 9.5.3.

Solve for $p$ in $-40=10-2(p-1)^2\text{.}$

Explanation

We isolate the squared binomial and then use the square root property.

\begin{align*} -40\amp=10-2(p-1)^2\\ -50\amp=-2(p-1)^2\\ 25\amp=(p-1)^2 \end{align*}

\begin{align*} p-1\amp=-5\amp\text{ or }\amp\amp p-1\amp=5\\ p\amp=-4\amp\text{ or }\amp\amp p\amp=6 \end{align*}

The solution set is $\{-4,6\}$

Let's check the solution $p=-4\text{:}$

\begin{align*} -40\amp=10-2(p-1)^2\\ -40\amp\stackrel{?}{=}10-2(\substitute{-4}-1)^2\\ -40\amp\stackrel{?}{=}10-2(-5)^2\\ -40\amp\stackrel{?}{=}10-2(25)\\ -40\amp\stackrel{\checkmark}{=}10-50 \end{align*}

The solution $p=-4$ is verified. Checking $p=6$ is left as an exercise.

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When we have a middle term in

$ax^2+bx+c=0\text{,}$ we cannot use the square root property. We look first to see if we can solve the equation by factoring. Here are some examples.

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Example 9.5.4.

Solve for $x$ in $x^2-4x-12=0\text{.}$

Explanation

The equation is already in standard form and we can factor the polynomial on the left side of the equation. We will factor it and then use the zero product property to solve the equation.

\begin{align*} x^2-4x-12\amp=0\\ (x-6)(x+2)\amp=0 \end{align*}

\begin{align*} x-6\amp=0\amp\text{ or }\amp\amp x+2\amp=0\\ x\amp=6\amp\text{ or }\amp\amp x\amp=-2 \end{align*}

The solution set is $\{-2,6\}\text{.}$

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Example 9.5.5.

Solve for $t$ in $2t^2-30t+28=0\text{.}$

Explanation

First, we factor out the common factor of $2\text{.}$ Then we can see that the polynomial is factorable. We solve it using the zero product property.

\begin{align*} 2t^2-30t+28\amp=0\\ 2(t^2-15t+14)\amp=0\\ 2(t-1)(t-14)\amp=0 \end{align*}

\begin{align*} t-1\amp=0\amp\text{ or }\amp\amp t-14\amp=0\\ t\amp=1\amp\text{ or }\amp\amp t\amp=14 \end{align*}

The solution set is $\{1,14\}\text{.}$

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If the equation is not in standard form, we must rewrite it before we can solve it. Let's look at the next example.

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Example 9.5.6.

Solve for $x$ in $(x+4)(x-3)=18\text{.}$

Explanation

We need to have one side equal to $0$ in order to use the zero product property, so we will multiply the left side and subtract $18$ from both sides.

\begin{align*} (x+4)(x-3)\amp=18\\ x^2+x-12\amp=18\\ x^2+x-12\subtractright{18}\amp=18\subtractright{18}\\ x^2+x-30\amp=0\\ (x+6)(x-5)\amp=0 \end{align*}

\begin{align*} x+6\amp=0\amp\text{ or }\amp\amp x-5\amp=0\\ x\amp=-6\amp\text{ or }\amp\amp x\amp=5 \end{align*}

The solution set is $\{-6,5\}\text{.}$

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When it's difficult or impossible to factor the trinomial in

$ax^2+bx+c=0\text{,}$ we have to resort to the Quadratic Formula:

$\begin{equation*} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{equation*}$

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Here is an example.

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Example 9.5.7.

Solve for $x$ in $x^2-10x+3=0\text{.}$

Explanation

We identify that $a=1\text{,}$ $b=-10$ and $c=3$ and substitute them into the Quadratic Formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(3)}}{2(1)}\\ x\amp=\frac{10\pm\sqrt{100-12}}{2}\\ x\amp=\frac{10\pm\sqrt{88}}{2}\\ x\amp=\frac{10\pm2\sqrt{22}}{2}\\ x\amp=5\pm\sqrt{22} \end{align*}

The solution set is $\left\{5-\sqrt{22}, 5+\sqrt{22}\right\}\text{.}$

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If a quadratic equation is not in standard form we need to rewrite it to identify the values of

$a\text{,}$

$b$ and

$c\text{.}$ Let's look at an example.

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Example 9.5.8.

Solve for $x$ in $-3x^2-1=-8x\text{.}$

Explanation

First, we convert the equation into standard form:

\begin{align*} -3x^2-1\amp=-8x\\ -3x^2+8x-1\amp=0 \end{align*}

Now we can identify that $a=-3\text{,}$ $b=8$ and $c=-1\text{.}$ We will substitute them into the Quadratic Formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-8\pm\sqrt{8^2-4(-3)(-1)}}{2(-3)}\\ x\amp=\frac{-8\pm\sqrt{64-12}}{-6}\\ x\amp=\frac{-8\pm\sqrt{52}}{-6}\\ x\amp=\frac{-8\pm2\sqrt{13}}{-6}\\ x\amp=\frac{4\pm\sqrt{13}}{3} \end{align*}

The solution set is $\left\{\frac{4-\sqrt{13}}{3}, \frac{4+\sqrt{13}}{3}\right\}$

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Also recall that if the radicand is negative, there is no real solution to the equation.

Solve the equation.

${x^{2}+13x} = {-40}$