Question

a quadratic equation is shown
$x^2 = -4$
select all true statements.
a $\square$ $x = -2$ is a solution to the equation.
b $\square$ $x = 2$ is a solution to the equation.
c $\square$ there are no real solutions to the equation.
d $\square$ there are 2 real solutions to the equation.
e $\square$ $x = \sqrt{-4}$ is a solution to the equation.
f $\square$ $x = \sqrt{-4}$ is an imaginary number.
$\square$ a
$\square$ b
$\square$ c
$\square$ d
$\square$ e
$\square$ f

Explanation:

Step1: Test real solution x=-2

Substitute $x=-2$ into $x^2=-4$: $(-2)^2=4
eq-4$

Step2: Test real solution x=2

Substitute $x=2$ into $x^2=-4$: $(2)^2=4
eq-4$

Step3: Analyze real solution existence

For real numbers, $x^2\geq0$, so no real $x$ satisfies $x^2=-4$.

Step4: Evaluate imaginary solution

$\sqrt{-4}=2i$, substitute into equation: $(2i)^2=4i^2=-4$, which matches. $\sqrt{-4}$ is an imaginary number.

Answer:

C. There are no real solutions to the equation.
E. $x = \sqrt{-4}$ is a solution to the equation.
F. $x = \sqrt{-4}$ is an imaginary number.