What are all real solutions to x^4 - 13x² + 36 = 0?

This equation is quadratic in form. Let $u=x^2$, so it becomes $u^2-13u+36=0$, which factors as $(u-4)(u-9)=0$. Then $u=4$ or $u=9$. Since $u=x^2$, we get $x^2=4$, so $x=\pm2$, and $x^2=9$, so $x=\pm3$. All four values work, so the answer is $x=\pm2,\pm3$, choice A. Choice B lists only the positive roots and misses that each square has two roots. Choice C ($\pm4,\pm9$) gives the values of $u$ with the wrong signs and never takes the square root. Choice D ($\pm6$) would make $x^2=36$, which does not satisfy the equation. Check $x=2$: $16-52+36=0$.