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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following equations by using the method of completing the square:
$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
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Answer

$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
$\Rightarrow\text{x}^2-\big(\sqrt2+1\big)\text{x}=-\sqrt2$
$\Rightarrow\text{x}^2-2\times\text{x}\times\Big(\frac{\sqrt2+1}{2}\Big)+\Big(\frac{\sqrt2+1}{2}\Big)^2\\=-\sqrt2+\Big(\frac{\sqrt2+1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{\sqrt2+1}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big[\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)\Big]^2=\frac{-4\sqrt2+2+1+2\sqrt2}{4}$
$=\frac{2-2\sqrt2+1}{4}=\Big(\frac{\sqrt2-1}{2}\Big)^2$
$\Rightarrow\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=\pm\Big(\frac{\sqrt2-1}{2}\Big)$ (Taking square root on both sides)
$\Rightarrow\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=\Big(\frac{\sqrt2-1}{2}\Big)$ or $\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=-\Big(\frac{\sqrt2-1}{2}\Big)$
$\Rightarrow\text{x}=\Big(\frac{\sqrt2+1}{2}\Big)+\Big(\frac{\sqrt2-1}{2}\Big)$ or $\text{x}=\Big(\frac{\sqrt2+1}{2}\Big)-\Big(\frac{\sqrt2-1}{2}\Big)$
$\Rightarrow\text{x}=\frac{2\sqrt2}{2}=\sqrt2$ or $\text{x}=\frac{2}2{}=1$
Hence, $\sqrt2$ and 1 are the roots of the given equation.

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