Solve the following quadratic equation by factorization:

$\sqrt{3}x^2-2\sqrt{2}x-2\sqrt3=0$

Given:

Given quadratic equation is $\sqrt{3}x^2-2\sqrt{2}x-2\sqrt3=0$.

To do:

We have to solve the given quadratic equation.

Solution:

$\sqrt{3}x^2-2\sqrt{2}x-2\sqrt3=0$

To factorise $\sqrt{3}x^2-2\sqrt{2}x-2\sqrt3=0$, we have to find two numbers $m$ and $n$ such that $m+n=-2\sqrt{2}$ and $mn=\sqrt{3}\times(-2\sqrt{3})=-2(\sqrt3)^2=-6$.

If $m=-3\sqrt2$ and $n=\sqrt2$, $m+n=-3\sqrt2+\sqrt2=-2\sqrt2$ and $mn=(-3\sqrt2)(\sqrt2)=-6$.

$\sqrt{3}x^2-3\sqrt{2}x+\sqrt{2}x-2\sqrt3=0$

$\sqrt{3}x(x-(\sqrt2)(\sqrt3))+\sqrt2(x-(\sqrt2)(\sqrt3))=0$

$(\sqrt{3}x+\sqrt2)(x-\sqrt6)=0$

$\sqrt{3}x+\sqrt2=0$ or $x-\sqrt6=0$

$\sqrt{3}x=-\sqrt2$ or $x=\sqrt6$

$x=\frac{-\sqrt2}{\sqrt3}$ or $x=\sqrt6$

$x=-\sqrt{\frac{2}{3}}$ or $x=\sqrt6$

The values of $x$ are $\sqrt{\frac{2}{3}}$ and $\sqrt6$.

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Updated on: 10-Oct-2022

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