Question

Solve each of the following quadratic equations:

$x^2+2\sqrt{2}x-6=0$

Answer 1

We write, $2\sqrt{2}x=3\sqrt{2}x-\sqrt{2}x$ as $x^2\times \left(-6\right)=-6x^2=3\sqrt{2}x\times \left(-\sqrt{2}x\right)$

$$\begin{gathered} \therefore x^2+2\sqrt{2}x-6=0 \\ \Rightarrow x^2+3\sqrt{2}x-\sqrt{2}x-6=0 \\ \Rightarrow x\left(x+3\sqrt{2}\right)-\sqrt{2}\left(x+3\sqrt{2}\right)=0 \\ \Rightarrow \left(x+3\sqrt{2}\right)\left(x-\sqrt{2}\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+3\sqrt{2}=0\mathrm{or}x-\sqrt{2}=0 \\ \Rightarrow x=-3\sqrt{2}\mathrm{or}x=\sqrt{2} \end{gathered}$$
Hence, the roots of the given equation are $-3\sqrt{2}$ and $\sqrt{2}$.