Question

Solve question no. 13 by completing the square method.

Answer 1

Question :-

Solve the equation x² - (√2 + 1)x + √2 = 0 by completing the square method.

Solution :-

${x}^{2} - ( \sqrt{2} - 1)x + \sqrt{2} = 0$

$\implies {x}^{2} - ( \sqrt{2} + 1)x = - \sqrt{2}$

$\implies {x}^{2} - 2(x) \bigg( \dfrac{ \sqrt{2} + 1 }{2} \bigg) = - \sqrt{2}$

Adding [ (√2 + 1)/2 ]² on both sides

$\implies {x}^{2} - 2(x) \bigg( \dfrac{ \sqrt{2} + 1 }{2} \bigg) + \bigg( \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2} = - \sqrt{2} + \bigg( \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2}$

$\implies \bigg( x - \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2} = - \sqrt{2} + \dfrac{( \sqrt{2} + 1 )^{2} }{2^{2} }$

[ Because a² - 2ab + b² = (a - b)² ]

$\implies \bigg( x - \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2} = - \sqrt{2} + \dfrac{( \sqrt{2})^{2} + 2( \sqrt{2} )(1) + {1}^{2} }{4 }$

$\implies \bigg( x - \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2} = \dfrac{ - 4 \sqrt{2} + ( \sqrt{2})^{2} + 2 \sqrt{2} + {1}^2 }{4 }$

$\implies \bigg( x - \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2} = \dfrac{ ( \sqrt{2})^{2} - 2 (\sqrt{2})(1) + {1}^2 }{4 }$

$\implies \bigg( x - \dfrac{ \sqrt{2} + 1 }{2} \bigg)^{2} = \dfrac{ ( \sqrt{2} - 1) ^2 }{4 }$

[ Because a² - 2ab + b² = (a - b)² ]

Taking square root on both sides

$\implies x - \dfrac{ \sqrt{2} + 1 }{2} = \pm \sqrt{ \dfrac{ ( \sqrt{2} - 1) ^2 }{4 } }$

$\implies x - \dfrac{ \sqrt{2} + 1 }{2} = \pm \dfrac{\sqrt{2} - 1 }{2 }$

$\implies x - \dfrac{ \sqrt{2} + 1 }{2} = \dfrac{\sqrt{2} - 1 }{2 } \ \ \ \ or \ \ \ x - \dfrac{ \sqrt{2} + 1}{2} = - \dfrac{ \sqrt{2} - 1}{2}$

$\implies x = \dfrac{ \sqrt{2} - 1 }{2} + \dfrac{\sqrt{2} + 1 }{2 } \ \ \ \ or \ \ \ x = \dfrac{ - ( \sqrt{2} - 1)}{2} + \dfrac{ \sqrt{2} + 1}{2}$

$\implies x = \dfrac{ \sqrt{2} - 1 + \sqrt{2} + 1 }{2} \ \ \ \ or \ \ \ x = \dfrac{ - \sqrt{2} + 1 + \sqrt{2} + 1}{2}$

$\implies x = \dfrac{2 \sqrt{2} }{2} \ \ \ \ or \ \ \ x = \dfrac{ 2}{2}$

$\implies x = \sqrt{2} \ \ \ \ or \ \ \ x = 1$

Hence, √2 and 1 are the roots of the equation.