Recognizing Pairs of Solutions

8.3.2 Recognizing Pairs of Solutions

8.3.2 • Recognizing Pairs of Solutions

Activity

A whole number that has two factors that are equal to each other is called a perfect square. Those equal factors that multiply to be the number are called the square root of a number. For instance, $3$ is a square root of $9$ because $3 \cdot 3 = 9 .$ .

SQUARE ROOT PROPERTY

If $x^{2} = k$ , then

$x = \sqrt{k} and x = - \sqrt{k} \to x = \pm \sqrt{k}$

Each of these equations has two solutions. Use your knowledge of square roots and the square root property. What are the solutions? Explain or show your reasoning.

$n^{2} + 4 = 404$

Compare your answer:

$n = 20$ and $n = - 20$

If $n^{2} + 4 = 404$ , then $n^{2} = 400$ . Using the square root property, $n = \pm \sqrt{400}$ . Therefore, $n = 20$ and $n = - 20$ .

$432 = 3 n^{2}$

Compare your answer:

$n = 12$ and $n = - 12$

Divide each side of the equation by 3. $n^{2} = 144$ Using the square root property, $n = \pm \sqrt{144}$ . Therefore, $n = 12$ and $n = - 12$ .

$\frac{n^{2}}{5} - 20 = 0$

Compare your answer:

$n = 10$ and $n = - 10$

Add 20 to each side of the equation. Then multiply each side of the equation by 5. $n^{2} = 100$ Using the square root property, $n = \pm \sqrt{100}$ . Therefore, $n = 10$ and $n = - 10$ .

$2 n^{2} - 7 = 65$

Compare your answer:

$n = 6$ and $n = - 6$

Add 7 to each side of the equation. Divide each side of the equation by 2. $n^{2} = 36$ Using the square root property, $n = \pm \sqrt{36}$ . Therefore, $n = 6$ and $n = - 6$ .

Are you ready for more?

Extending Your Thinking

How many solutions does the equation $(x - 3) (x + 1) (x + 5) = 0$ have? What are the solutions?

Compare your answer:

Three solutions: $x = 3$ , $x = - 1$ , and $x = - 5$ .

How many solutions does the equation $(x - 2) (x - 7) (x - 2) = 0$ have? What are the solutions?

Compare your answer:

Two solutions: $x = 2$ and $x = 7$ .

Write a new equation that has 10 solutions.

Compare your answer:

There are many different possible answers. For example:

$(x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) (x - 7) (x - 8) (x - 9) (x - 10) = 0$

Self Check

Solve for $p$ . Find all solutions.

$3p^2+8 =251$

$p=9$ and $p=-9$
$p=11$
$p=11$ and $p=-11$
$p=9$

Additional Resources

Using the Square Root Property to Solve Quadratic Equations

Let’s look again at the square root property.

SQUARE ROOT PROPERTY

If $x^{2} = k$ , then

$x = \sqrt{k} and x = - \sqrt{k} \to x = \pm \sqrt{k}$

Notice that the square root property gives two solutions to an equation of the form $x^{2} = k$ , the principal square root of $k$ and its opposite. We could also write the solution as $x = \pm \sqrt{k}$ . We read this as $x$ equals positive and negative the square root of $k$ .

Example 1

Let’s solve $x^{2} = 9$ using the square root property.

$x^{2} = 9$

$x = \pm \sqrt{9}$

$x = \pm 3$

So, $x = 3$ and $x = - 3$ .

If the number in the radical is not a perfect square, leave that part inside the radical. Let’s solve another example.

Example 2

Solve $x^{2} - 50 = 0$ .

Step 1 - Isolate the quadratic term and make its coefficient 1.

To do this, add 50 to both sides of the equation to get $x^{2}$ by itself.

$x^{2} - 50 = 0$

$x^{2} = 50$

Step 2 - Use the square root property.

Remember to write the symbol.

$x = \pm \sqrt{50}$

Step 3 - Simplify the radical by pulling out the factors that are perfect squares.

$x = \pm \sqrt{25} \cdot \sqrt{2}$

$x = \pm 5 \sqrt{2}$

$x = 5 \sqrt{2}$ , $x = - 5 \sqrt{2}$

Step 4 - Check the solutions.

$\begin{aligned} x^{2} - 50 = 0 \\ (5 \sqrt{2})^{2} - 50 \overset{?}{=} 0 \\ 25 \cdot 2 - 50 \overset{?}{=} 0 \\ 0 = 0 ✓ \\ x^{2} - 50 = 0 \\ (- 5 \sqrt{2})^{2} - 50 \overset{?}{=} 0 \\ 25 \cdot 2 - 50 \overset{?}{=} 0 \\ 0 = 0 ✓ \end{aligned}$

The steps for solving a quadratic equation using the square root property are listed here.

How to solve a quadratic equation using the square root property.

Step 1 - Isolate the quadratic term and make its coefficient 1.

Step 2 - Use the square root property.

Step 3 - Simplify the radical.

Step 4 - Check the solutions.

Before using the square root property in Step 2, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by the coefficient 3 before using the square root property.

Example 3

Solve $3 z^{2} = 108$ .

Step 1 - Isolate the quadratic term and make its coefficient 1.

When the quadratic term is isolated, divide by 3 to make its coefficient 1. Simplify.

$3 z^{2} = 108$

$\frac{3 z^{2}}{3} = \frac{108}{3}$

$z^{2} = 36$

Step 2 - Use the square root property.

$z = \pm \sqrt{36}$

Step 3 - Simplify the radical.

Rewrite to show two solutions.

$z = \pm 6$

$z = 6$ , $z = - 6$

Step 4 - Check the solutions.

$\begin{matrix} 3 z^{2} = 108 & 3 z^{2} = 108 \\ 3 (6)^{2} \overset{?}{=} 108 & 3 (- 6)^{2} \overset{?}{=} 108 \\ 3 (36) \overset{?}{=} 108 & 3 (36) \overset{?}{=} 108 \\ 108 = 108 ✓ & 108 = 108 ✓ \end{matrix}$

Try it

Using the Square Root Property to Solve Quadratic Equations

Solve the following quadratic equations using the square root property.

1. $2 n^{2} = 98$

2. $\frac{n^{2}}{2} - 30 = 68$

Here is how to solve these quadratic equations using the square root property:

$2 n^{2} = 98 \frac{2 n^{2}}{2} = \frac{98}{2} n^{2} = 49 n = \pm \sqrt{49} n = 7 and n = - 7$

$\frac{n^{2}}{2} - 30 = 68 \frac{n^{2}}{2} = 98 n^{2} = 196 n = \pm \sqrt{196} n = 14 and n = - 14$