Question

Please solve this on a paper will be given brainliest ​

Answer 1

Answer:

$x=2,\\x=-2,\\x=\sqrt{3},\\x=-\sqrt{3}$

Step-by-step explanation:

This is a quadratic in disguise. Let $n=x^2$. We can rewrite the problem as a quadratic to solve:

$n^2-7n+12=0,\\(n-3)(n-4)=0,\\n=3,\: n=4$.

Now plug in these values of $n$ to solve:

$\begin{cases}x^2=4, \:x=\pm 2\\\\x^2=3, x=\pm\sqrt{3}\end{cases}$.

Therefore, our solutions are $x=\pm 2, \:x=\pm\sqrt{3}$.