solve (5-b)^2 = y for b — step-by-step solution

1) Let's solve in

ℝ

{(5 - b)}^{2} = y

A square being never negative, we need

y \geq 0

; the two numbers whose square is

y

are then

- \sqrt{y}

and

\sqrt{y}

5 - b = - \sqrt{y}

ⓘ We subtract

5

from both sides

- b = - \sqrt{y} - 5

ⓘ We gather the terms in

b

on the right and the other terms on the left

\sqrt{y} + 5 = b

Solution found:

b = \sqrt{y} + 5

5 - b = \sqrt{y}

ⓘ We subtract

5

from both sides

- b = \sqrt{y} - 5

ⓘ We gather the terms in

b

on the right and the other terms on the left

- \sqrt{y} + 5 = b

Solution found:

b = 5 - \sqrt{y}

2) Solving for y:

y \geq 0

Solution found:

y \geq 0

▶Solution:

{b = {5 - \sqrt{y}, \sqrt{y} + 5}, y \geq 0}