Sure! Let's solve each of these equations step-by-step:
### 1) [tex]\( 2n^2 = 50 \)[/tex]
1. Divide both sides by 2 to isolate the square term:
[tex]\[
n^2 = \frac{50}{2} = 25
\][/tex]
2. Take the square root of both sides to solve for [tex]\( n \)[/tex]:
[tex]\[
n = \pm \sqrt{25}
\][/tex]
3. Simplify the square root:
[tex]\[
n = \pm 5
\][/tex]
So, the solutions to [tex]\( 2n^2 = 50 \)[/tex] are [tex]\( n = -5 \)[/tex] and [tex]\( n = 5 \)[/tex].
### 2) [tex]\( 4x^2 - 225 = 0 \)[/tex]
1. Add 225 to both sides to isolate the square term:
[tex]\[
4x^2 = 225
\][/tex]
2. Divide both sides by 4:
[tex]\[
x^2 = \frac{225}{4}
\][/tex]
3. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x = \pm \sqrt{\frac{225}{4}}
\][/tex]
4. Simplify the square root:
[tex]\[
x = \pm \frac{\sqrt{225}}{\sqrt{4}} = \pm \frac{15}{2}
\][/tex]
So, the solutions to [tex]\( 4x^2 - 225 = 0 \)[/tex] are [tex]\( x = -\frac{15}{2} \)[/tex] and [tex]\( x = \frac{15}{2} \)[/tex].
### 3) [tex]\( (x - 4)^2 = 169 \)[/tex]
1. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x - 4 = \pm \sqrt{169}
\][/tex]
2. Simplify the square root:
[tex]\[
x - 4 = \pm 13
\][/tex]
3. Solve for [tex]\( x \)[/tex] by adding 4 to both sides of each equation:
[tex]\[
x = 4 + 13 \quad \text{or} \quad x = 4 - 13
\][/tex]
4. Simplify:
[tex]\[
x = 17 \quad \text{or} \quad x = -9
\][/tex]
So, the solutions to [tex]\( (x - 4)^2 = 169 \)[/tex] are [tex]\( x = 17 \)[/tex] and [tex]\( x = -9 \)[/tex].
### Summary
1. The solutions for [tex]\( 2n^2 = 50 \)[/tex] are [tex]\( n = -5 \)[/tex] and [tex]\( n = 5 \)[/tex].
2. The solutions for [tex]\( 4x^2 - 225 = 0 \)[/tex] are [tex]\( x = -\frac{15}{2} \)[/tex] and [tex]\( x = \frac{15}{2} \)[/tex].
3. The solutions for [tex]\( (x - 4)^2 = 169 \)[/tex] are [tex]\( x = 17 \)[/tex] and [tex]\( x = -9 \)[/tex].
Thus, the final results are:
[tex]\[ [-5, 5], \left[-\frac{15}{2}, \frac{15}{2}\right], [-9, 17] \][/tex].
