Question

solve each quadratic equation using square roots.
1.) $x^2 = 324$
2.) $x^2 - 81 = 0$
3.) $5x^2 - 180 = 0$

4.) $3x^2 - 100 = 332$
5.) $\frac{1}{2}x^2 - 5 = 5$
6.) $(3x + 1)^2 = 36$

7.) $2(x + 3)^2 = 8$
8.) $2(x + 4)^2 = 10$
9.) $(x + 1)^2 = - 16$

Explanation:

1. Solve \( x^2 = 324 \)

Step1: Take square root of both sides

To solve for \( x \), we take the square root of both sides of the equation \( x^2 = 324 \). Remember that when we take the square root of a number, we get both a positive and a negative solution because \( (\pm a)^2=a^2 \). So we have:
\( x=\pm\sqrt{324} \)

Step2: Simplify the square root

We know that \( 18\times18 = 324 \) and \( (- 18)\times(-18)=324 \), so \( \sqrt{324} = 18 \). Therefore:
\( x=\pm18 \)

Step1: Isolate \( x^2 \)

First, we add 81 to both sides of the equation to isolate \( x^2 \). So:
\( x^2-81 + 81=0 + 81 \)
\( x^2=81 \)

Step2: Take square root of both sides

Now we take the square root of both sides. Since \( (\pm9)^2 = 81 \), we have:
\( x=\pm\sqrt{81}=\pm9 \)

Step1: Isolate \( x^2 \)

First, we add 180 to both sides of the equation:
\( 5x^2-180 + 180=0 + 180 \)
\( 5x^2=180 \)
Then we divide both sides by 5 to solve for \( x^2 \):
\( \frac{5x^2}{5}=\frac{180}{5} \)
\( x^2 = 36 \)

Step2: Take square root of both sides

Taking the square root of both sides, we get:
\( x=\pm\sqrt{36}=\pm6 \)

Answer:

\( x = 18 \) or \( x=- 18 \)

2. Solve \( x^2-81 = 0 \)