8.EE.A.2 Tutor Intake -- Square Roots and Cube Roots

$\sqrt{25} = 5$. Only the equation $x^2 = 25$ has two solutions, $x = 5$ and $x = -5$.

$\sqrt{25} = \pm 5$, because both $5^2$ and $(-5)^2$ equal 25.

$\sqrt{25} = -5$, since the radical symbol always gives the negative root.

$\sqrt{25} = 5$ and $\sqrt{25} = -5$ are both correct at the same time.

One solution, $x = 2$, because a negative number cubed stays negative, so $-2$ does not work.

Two solutions, $x = \pm 2$, using the same pattern as square root equations.

One solution, $x = -2$, because cube roots are always negative.

Two solutions, $x = 2$ and $x = 4$.

1.414 is only an approximation, not the exact value; the true decimal expansion of $\sqrt{2}$ goes on forever without repeating, which is what makes it irrational.

Nothing is wrong — since the calculator shows a terminating decimal, $\sqrt{2}$ is in fact rational.

The reasoning is wrong because $\sqrt{2}$ is actually a perfect square.

The reasoning is wrong because calculators cannot compute square roots accurately.