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

$x^2 - (\sqrt2+1)x + \sqrt2 = 0$

Given:

Given quadratic equation is $x^2 - (\sqrt2+1)x + \sqrt2 = 0$.

To do:

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

Solution:

$x^2 - (\sqrt2+1)x + \sqrt2 = 0$

$x^2-2\times \frac{1}{2} \times (\sqrt2+1)x =-\sqrt2$

$x^2-2(\frac{\sqrt2+1}{2})x=-\sqrt2$

Adding $(\frac{\sqrt2+1}{2})^2$ on both sides completes the square. Therefore,

$x^2-2(\frac{\sqrt2+1}{2})x+(\frac{\sqrt2+1}{2})^2=-\sqrt2+(\frac{\sqrt2+1}{2})^2$

$(x-\frac{\sqrt2+1}{2})^2=-\sqrt2+\frac{2+1+2\sqrt2}{4}$      (Since $(a-b)^2=a^2-2ab+b^2$)

$(x-\frac{\sqrt2+1}{2})^2=\frac{3+2\sqrt2-\sqrt2\times4}{4}$

$(x-\frac{\sqrt2+1}{2})^2=\frac{3+2\sqrt2-4\sqrt2}{4}$

$(x-\frac{\sqrt2+1}{2})^2=\frac{3-2\sqrt2}{4}$

$x-\frac{\sqrt2+1}{2}=\pm \sqrt{\frac{(\sqrt2)^2+(1)^2-2(1)\sqrt2}{(2)^2}}$

$x-\frac{\sqrt2+1}{2}=\pm \sqrt{\frac{(\sqrt2-1)^2}{(2)^2}}$

$x-\frac{\sqrt2+1}{2}=\pm \sqrt{(\frac{(\sqrt2-1}{2})^2}$

$x-\frac{\sqrt2+1}{2}=\pm \frac{\sqrt2-1}{2}$

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

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

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

$x=\sqrt2$ or $x=1$

The values of $x$ are $1$ and $\sqrt{2}$.

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Updated on: 10-Oct-2022

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