Question

Which equation are true for x=-2 and x=2 check all that apply

Answer 1

1 it is true
${x}^{2} - 4 = 0 \\ {x}^{2} = + 4 \\ x = \sqrt{4} = 2$
2 it's not true
${x}^{2} = - 4$
3 it's not true
$3 {x}^{2} + 12 = 0 \\ 3 {x}^{2} = - 12 \\ {x}^{2} = \frac{ - 12}{3} = - 4$
4 it's true
$4 {x}^{2} = 16 \\ {x}^{2} = \frac{16}{4} = 4 \\ x = \sqrt{4} = 2$
5- it's true
${2(x - 2)}^{2} = 0 \\ {(x - 2)}^{2} = 0 \\ x - 2 = 0 \\ x = 2$

Answer 2

Answer:

A. $x^2-4=0$

D. $4x^2=16$

Step-by-step explanation:

To check which of the given equations have solution $x=-2\text{ and }x=2$, we will solve our given equations one by one.

A. $x^2-4=0$

$x^2-4+4=0+4$

$x^2=4$

$\sqrt{x^2}=\pm\sqrt{4}$

$x=\pm 2$  

$x=-2\text{ or }x=2$  

Therefore, option A is the correct choice.

B. $x^2=-4$

To solve our given equation, we need to take square root of both sides of equation. Since square root is not defined for negative numbers, therefore, option B is not a correct choice.

C. $3x^2+12=0$

$3x^2=-12$

$x^2=-4$

To solve our given equation, we need to take square root of both sides of equation. Since square root is not defined for negative numbers, therefore, option C is not a correct choice.

D. $4x^2=16$

$x^2=4$

$\sqrt{x^2}=\pm\sqrt{4}$

$x=\pm 2$  

$x=-2\text{ or }x=2$  

Therefore, option D is the correct choice.

E. $2(x-2)^2=0$

$(x-2)^2=0$  

$\sqrt{(x-2)^2}=\sqrt{0}$  

$x-2=0$  

$x=2$  

Therefore, option E is not a correct choice.