Question

Solve each of the following quadratic equations:

$\sqrt{7}{x}^2-6x-13\sqrt{7}=0$

Answer 1

We write, $-6x=7x-13x$ as $\sqrt{7}{x}^2\times \left(-13\sqrt{7}\right)=-91x^2=7x\times \left(-13x\right)$

$$\begin{gathered} \therefore \sqrt{7}{x}^2-6x-13\sqrt{7}=0 \\ \Rightarrow \sqrt{7}{x}^2+7x-13x-13\sqrt{7}=0 \\ \Rightarrow \sqrt{7}x\left(x+\sqrt{7}\right)-13\left(x+\sqrt{7}\right)=0 \\ \Rightarrow \left(x+\sqrt{7}\right)\left(\sqrt{7}x-13\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x+\sqrt{7}=0\mathrm{or}\sqrt{7}x-13=0 \\ \Rightarrow x=-\sqrt{7}\mathrm{or}x=\frac{13}{\sqrt{7}}=\frac{13\sqrt{7}}{7} \end{gathered}$$
Hence, the roots of the given equation are $-\sqrt{7}$ and $\frac{13\sqrt{7}}{7}$.