2. The equations $x^{2}-a=0$ and $3 x^{4}-48=0$ have the same real solutions. What is the value of $a$ ?
Solution 1: The left hand side of the equation $3 x^{4}-48=0$ can be factored as $3\left(x^{4}-16\right)=$ $3\left(x^{2}-4\right)\left(x^{2}+4\right)=3(x-2)(x+2)\left(x^{2}+4\right)$.
Thus, the real solutions of the equation $3 x^{4}-48=0$ are $x= \pm 2$, so $x^{2}=4$, and $a=4$.
Solution 2: The equation $3 x^{4}-48=0$ can be factored as $3\left(x^{2}-4\right)\left(x^{2}+4\right)=0$. Since this must have the same real solutions as $x^{2}-a$ we must have $x^{2}-a=x^{2}-4$ and so $a=4$.