Question

solve
${x}^{2} - ( \sqrt{2} + 1)x + \sqrt{2} = 0$
by completing the square​

Answer 1

GIVEN :–

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

TO FIND :–

• Roots of equation by completing perfect square ?

SOLUTION :–

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

• We should write this as –

$\\\implies\rm{x}^{2} - ( \sqrt{2} + 1)x + \sqrt{2} + \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} - \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} = 0$

$\\\implies\rm{x}^{2} - ( \sqrt{2} + 1)x + \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} + \sqrt{2} - \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} = 0$

$\\\implies\rm{x}^{2} - ( \sqrt{2} + 1)x + \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} + \sqrt{2} - \bigg( \dfrac{2+1 + 2 \sqrt{2} }{4}\bigg) = 0$

$\\\implies\rm{x}^{2} - ( \sqrt{2} + 1)x + \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} + \sqrt{2} - \bigg( \dfrac{ 3+ 2 \sqrt{2} }{4}\bigg) = 0$

$\\\implies\rm{x}^{2} - ( \sqrt{2} + 1)x + \bigg( \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} + \dfrac{1}{ \sqrt{2} } - \dfrac{3}{4} = 0$

• Using identity –

$\\\longrightarrow \: \: \rm{a}^{2} -2ab+ {b}^{2} = {(a - b)}^{2}$

• So that –

$\\\implies\rm\bigg(x - \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} = \dfrac{3}{4} - \dfrac{1}{ \sqrt{2} }$

$\\\implies\rm\bigg(x - \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} = \dfrac{3}{4} - \dfrac{2 \sqrt{2} }{4}$

$\\\implies\rm\bigg(x - \dfrac{ \sqrt{2} + 1}{2}\bigg)^{2} = \dfrac{3 - 2 \sqrt{2} }{4}$

$\\\implies\rm x - \dfrac{ \sqrt{2} + 1}{2} = \sqrt{ \dfrac{3 - 2 \sqrt{2} }{4}}$

$\\\implies\rm x - \dfrac{ \sqrt{2} + 1}{2} = \pm \dfrac{ \sqrt{3 - 2 \sqrt{2}}}{2}$

$\\\implies\rm x = \dfrac{ \sqrt{2} + 1}{2} \pm \dfrac{ \sqrt{3 - 2 \sqrt{2}}}{2}$

$\\ \large\implies{ \boxed{\rm x = \dfrac{(\sqrt{2} + 1) \pm(\sqrt{3 - 2 \sqrt{2}})}{2}}}$

• Hence , The roots are $\rm \dfrac{(\sqrt{2} + 1)+(\sqrt{3 - 2 \sqrt{2}})}{2}\:\: and\:\: \dfrac{(\sqrt{2} + 1)-(\sqrt{3 - 2 \sqrt{2}})}{2}$

Answer 2

Method 1

Given :

x²-(√2+1)x +√2 = 0

To Find :

• by completing the square

Solution :

Factors of +√2 are -√2 and -1.

So x²-(√2+1)x +√2 = 0 can be written as:

x² -√2x -x +√2 = 0

x (x -√2) -1 (x -√2) = 0

(x - 1) (x -√2) = 0

So the factors are (x - 1) and (x -√2).

If x - 1 = 0, then x = +1

If x -√2 = 0, then x = +√2

Hence solved.

X = 1, √2

___________________

Method 2

Given :

• a = 1 , b= -(√2+1) and c= √2

To Find

• by completing the square

Solution :

D = b²- 4ac

Substitute all values :

= ( - 2 - 1 - 2√2) - 4 × 1 ×√2

= (- 3 - 2√2) - 4√2

= - 3 - 2√2 - 4√2

= - 3 - 6√2

= - 3(1 + 2√2)

x = - b ± √D / 2a

Substitute values :

= (√2 + 1) ± √- 3(1 + 2√2) /2