Solve by taking Square Roots?

Question

x^2-32=0

(t-1)^2-15=0

Answer 1

PROBLEM 1:

x² - 32 = 0

Add 32 to both sides:
x² = 32

Take the square root of both sides:
x = ±√32
x = ±√(16)(2)
x = ±4√2

PROBLEM 2:

(t -1)² - 15 = 0

Add 15 to both sides:
(t -1)² = 15

Take the square root of both sides:
t -1 = ±√15

Add 1 to both sides:
t = 1 ± √15

P.S. Don't forget to include both the positive and negative square roots...

Answer 2

x^2 = 32
=> x = +/-sqrt(32)
= +/-sqrt(16*2)
= +/-sqrt(16)sqrt(2)
= +/-4*sqrt(2)

(t-1)^2 = 15
t-1 = +/-sqrt(15)
t = 1 +/- sqrt(15)

Answer 3

x^2-32=0
x^2=32
x=+/- sq.rt(32)

(t-1)^2-15=0
(t-1)^2=15
t-1=+/- sq.rt(15)
t= 1 +/- sq.rt(15)

Answer 4

x^2 = 32
x=+-sqrt(32)

(t-1)^2 = 15
t-1=+-sqrt(15)
t=sqrt(15)+1
t=-sqrt(15)+1

Answer 5

x^2-32=0

x^2=32

Take the square root of both #'s and we get...

x=5.656