Answer:
A. [tex]x^2-4=0[/tex]
D. [tex]4x^2=16[/tex]
Step-by-step explanation:
To check which of the given equations have solution [tex]x=-2\text{ and }x=2[/tex], we will solve our given equations one by one.
A. [tex]x^2-4=0[/tex]
[tex]x^2-4+4=0+4[/tex]
[tex]x^2=4[/tex]
[tex]\sqrt{x^2}=\pm\sqrt{4}[/tex]
[tex]x=\pm 2[/tex]
[tex]x=-2\text{ or }x=2[/tex]
Therefore, option A is the correct choice.
B. [tex]x^2=-4[/tex]
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. [tex]3x^2+12=0[/tex]
[tex]3x^2=-12[/tex]
[tex]x^2=-4[/tex]
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. [tex]4x^2=16[/tex]
[tex]x^2=4[/tex]
[tex]\sqrt{x^2}=\pm\sqrt{4}[/tex]
[tex]x=\pm 2[/tex]
[tex]x=-2\text{ or }x=2[/tex]
Therefore, option D is the correct choice.
E. [tex]2(x-2)^2=0[/tex]
[tex](x-2)^2=0[/tex]
[tex]\sqrt{(x-2)^2}=\sqrt{0}[/tex]
[tex]x-2=0[/tex]
[tex]x=2[/tex]
Therefore, option E is not a correct choice.
