Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

$\sqrt2x^2 -3x - 2\sqrt2 = 0$

Given:

Given quadratic equation is $\sqrt2x^2 -3x - 2\sqrt2 = 0$.

To do:

We have to find the roots of the given quadratic equation.

Solution:

$\sqrt2x^2 -3x - 2\sqrt2 = 0$

$\sqrt2(x^2 -\frac{3}{\sqrt2} x -\frac{2\sqrt2}{\sqrt2})=0$  

$x^2-\frac{3}{\sqrt2}x-2=0$

$x^2-2\times \frac{1}{2} \times \frac{3}{\sqrt2} x =2$

$x^2-2(\frac{3}{2\sqrt2})x=2$

Adding $(\frac{3}{2\sqrt2})^2$ on both sides completes the square. Therefore,

$x^2-2(\frac{3}{2\sqrt2})x+(\frac{3}{2\sqrt2})^2=2+(\frac{3}{2\sqrt2})^2$

$(x-\frac{3}{2\sqrt2})^2=2+\frac{9}{8}$      (Since $(a-b)^2=a^2-2ab+b^2$)

$(x-\frac{3}{2\sqrt2})^2=\frac{9+2\times8}{8}$

$(x-\frac{3}{2\sqrt2})^2=\frac{9+16}{8}$

$x-\frac{3}{2\sqrt2}=\pm \sqrt{\frac{25}{8}}$

$x-\frac{3}{2\sqrt2}=\pm \frac{5}{2\sqrt2}$

$x=\frac{5}{2\sqrt2}+\frac{3}{2\sqrt2}$ or $x=\frac{3}{2\sqrt2}-\frac{5}{2\sqrt2}$

$x=\frac{5+3}{2\sqrt2}$ or $x=\frac{3-5}{2\sqrt2}$

$x=\frac{8}{2\sqrt2}$ or $x=\frac{-2}{2\sqrt2}$

$x=2\sqrt2$ or $x=-\frac{1}{\sqrt2}$

The values of $x$ are $2\sqrt2$ and $-\frac{1}{\sqrt2}$.

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Updated on: 10-Oct-2022

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