Exercise 10

Correct: $$\begin{align*} \sqrt{8x^2+2}=4x & \Rightarrow 8x^2+2=16x^2\\ & \Leftrightarrow 8x^2=2\\ & \Leftrightarrow x^2=\frac{1}{4}\\ & \Leftrightarrow x=\frac{1}{2} \mbox{ or } x=-\frac{1}{2}.\\ \end{align*}$$

In the first step we used $\Rightarrow$ instead of $\Leftrightarrow$. (The reason is that $a^2=b^2$ implies that $a=b$ or $a=-b$. If $x^2=4$ you only know that $x=-2$ or $x=2$.) Hence, we have to verify whether both $x=\frac{1}{2}$ and $x=-\frac{1}{2}$ are solutions of the original equation:

$\sqrt{8(\frac{1}{2})^2+2}=2=2=4\cdot \frac{1}{2}$, but
$\sqrt{8(-\frac{1}{2})^2+2}=2\neq-2=4\cdot -\frac{1}{2}$.

Hence, only $x=\frac{1}{2}$ is a solution of the equation.

Wrong: $\sqrt{8(-\frac{1}{2})^2+2}=2\neq-2=4\cdot -\frac{1}{2}$.

Try again.

Wrong: $\sqrt{\frac{1}{4}}\neq \frac{1}{16}$.

Try again.

Wrong: The correct answer is given.

Try again.