solve using the square root property: (x+7)^2=81?

(x+7)^2 = 81 = 9^2
Taking sqrt of both sides

(x+7) = +/- 9

or x+7 = 9 and x+7 = -9

x+7=9 gives x = 2 and x+7 = -9 gives x = -16

So both the values of x i.e. 2 and -16 satisfy the given equation.

(x+7)² = 81
x + 7 = ± 9

for x + 7 = 9
x =2

for x + 7 = - 9
x = - 16

(x + 7)^2 = 81. Square root both sides.
x + 7 = 9. x + 7 = -9.
x = 2. x = -16.

Hope this helps.

residences of sq. Roots PRODUCT supplies OF sq. ROOTS For all beneficial genuine numbers a and b, that's, the sq. root of the product is the same because of the fact the made from the sq. roots. QUOTIENT supplies OF sq. ROOTS For all beneficial genuine numbers a and b, b ? 0: The sq. root of the quotient is the same because of the fact the quotient of the sq. roots.

x^2+14x+49=81
x^2+14x-32=0

14^2-4(-32)=
196+128=
324

x1=-16
x2=2

(x + 7)^2 = 81
(x + 7)(x + 7) = 81
x^2 + 7x + 7x + 49 = 81
x^2 + 14x = 81 - 49
x^2 + 14x = 32
x^2 + 14x - 32 = 0
(x + 16)(x - 2) = 0

x + 16 = 0
x = -16

x - 2 = 0
x = 2

Take the square root on both sides.

If you write both sides as multiplication:

(x+7)*(x+7) = 9*9

taking the square root of a square is easy.

√(a*a) = a

However, remember that a square root can be positive or negative.

If the original number is a^2, then it could be factored as (-a)*(-a); so that -a is also a square root of a^2

You'll get
(x+7) = +/- 9

You'll need to solve using +9 (one answer) and then again, using -9 (another answer)

(x + 7)^2 = (x + 7) (x + 7)

x^2 + 14x + 49 = 81
x^2 + 14x - 32 = 0
(x + 16) (x - 2) = 0
x = -16 or 2