Solve $\sqrt{8x^2+2}=4x$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x=\frac{1}{2}$ or $x=-\frac{1}{2}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$x=\frac{1}{16}$ or $x=-\frac{1}{16}$
Antwoord 3 correct
Fout
Antwoord 4 optie
None of the other answers is correct.
Antwoord 4 correct
Fout
Antwoord 1 optie
$x=\frac{1}{2}$
Antwoord 1 feedback
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.
\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.
Antwoord 2 feedback
Wrong: $\sqrt{8(-\frac{1}{2})^2+2}=2\neq-2=4\cdot -\frac{1}{2}$.
Try again.
Try again.
Antwoord 3 feedback
Wrong: $\sqrt{\frac{1}{4}}\neq \frac{1}{16}$.
Try again.
Try again.
Antwoord 4 feedback
Wrong: The correct answer is given.
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Try again.