Question

Solving Square Roots Worksheet (x - k)^2 : Part 1

1. 2(x + 7)^2 = 16

2. (x - 3)^2 = -12

3. -5(n - 3)^2 = 10

Answer 1

Answer:

1. x = +/- 2$\sqrt{2}$ - 7

2. x = $3$ +/- $2i\sqrt{3}$

3. n = $2$ +/- $i\sqrt{2}$

Step-by-step explanation:

1. Divide both sides by 2: (x + 7)^2 = 8

Square root both sides: x + 7 = +/- 2$\sqrt{2}$

Subtract 7 from both sides: x = +/- 2$\sqrt{2}$ - 7

2. Square root both sides: x - 3 = $\sqrt{-12}$

Since there is a negative inside the radical, we need to have an imaginary number: $i=\sqrt{-1}$ . So, $\sqrt{-12} =i\sqrt{12} =2i\sqrt{3}$

Add 3 to both sides: x = $3$ +/- $2i\sqrt{3}$

3. Divide by -5 from both sides: (n - 2)^2 = -2

Square root both sides: n - 2 = $\sqrt{-2}$

Again, we have to use i: $n-2=\sqrt{-2} =i\sqrt{2}$

Add 2 to both sides: n = $2$ +/- $i\sqrt{2}$

Hope this helps!

Answer 2

Answer:

1. x = 2sqrt(2) - 7, -2sqrt(2) - 7

2. No real solutions

x = 3 + 2sqrt(3) i, 3 - 2sqrt(3) i

3. No real solutions

n = 3 + sqrt(2) i, 3 - sqrt(2) i

Step-by-step explanation:

1. 2(x + 7)² = 16

(x + 7)² = 8

x + 7 = +/- sqrt(8) = +/- 2sqrt(2(

x = 2sqrt(2) - 7, -2sqrt(2) - 7

2. (x - 3)² = -12

A perfect square can never be negative for real values of x

(x - 3) = +/- i × sqrt(12)

x - 3 = +/- i × 2sqrt(3)

x = 3 +/- i × 2sqrt(3)

3. -5(n - 3)² = 10

(n - 3)² = -2

A perfect square can never be negative for real values of x

n - 3 = +/- i × sqrt(2)

n = 3 +/- i × sqrt(2)