Question

Which are solutions of the equation x^2-16=0? Check all that apply.

Answer 1

Answer:

+4, -4

Step-by-step explanation:

There are only two solutions to this equation.

=> x^2 - 16 = 0

=> x^2 - 4^2 = 0  

=> (x+4) (x-4) = 0  {According to a^2 - b^2 = (a+b) (a-b)}

=> x = 4, -4

Answer 2

Answer:

x = 4

x = - 4

Step-by-step explanation:

Method 1: Quadratic Formula

Ignore the A before the ±, it wouldn't let me type it correctly.

$x=\frac{-b±\sqrt{b^{2} -4ac} }{2a}$

x² - 16 = 0

a = 1

b = 0

c = - 16

$x=\frac{-(0)±\sqrt{0^{2} -4((1)(-16))} }{2(1)}$

$x=\frac{0±\sqrt{0 -4((1)(-16))} }{2(1)}$

$x=\frac{0±\sqrt{0 +64} }{2(1)}$

$x=\frac{0±\sqrt{64} }{2}$

$x=\frac{0±8 }{2}$

Two separate equations

One must have a + (positive) and the other will have a - (negative).

$x=\frac{0+8 }{2}$

$x=\frac{8 }{2}$

x = 8 ÷ 2

x = 4

$x=\frac{-8 }{2}$

x = - 8 ÷ 2

x = - 4

Method 2: Factoring

x² - 16 = 0

(x - 4)(x + 4) = 0

Two separate equations

x - 4 = 0

x + 4 = 0

x - 4 = 0

x - 4 + 4 = 0 + 4

x = 4

x + 4 = 0

x + 4 - 4 = 0 - 4

x = - 4