Question

solve this this is a question of quadratic equations​

Answer 1

Question :-

Solve the equation x² - (√3 + 1)x + √3 = 0 by method of completing the square

Solution :-

x² - (√3 + 1)x + √3 = 0

⇒ x² - (√3 + 1)x = - √3

Adding [(√3 + 1)/2]² on both sides

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

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

Taking square root on both sides

$\\ \\ \sf \implies x - \dfrac{ \sqrt{3} + 1 }{2} = \pm \bigg( \dfrac{ \sqrt{3} - 1 }{2} \bigg) \\ \\ \\ \sf \implies x - \dfrac{ \sqrt{3} + 1}{2} = \dfrac{ \sqrt{3} - 1}{2} \: \: or \: \: x - \dfrac{ \sqrt{3} +1 }{2} = - \bigg(\dfrac{ \sqrt{3} - 1 }{2} \bigg) \\ \\ \\ \sf \implies x = \dfrac{ \sqrt{3} - 1}{2} + \dfrac{ \sqrt{3} + 1 }{2} \: \: or \: \: x = \dfrac{ \sqrt{3} + 1}{2} - \dfrac{ \sqrt{3} - 1 }{2} \\ \\ \\ \sf \implies x = \dfrac{ \sqrt{3} - 1 + \sqrt{3} + 1}{2} \: \: or \: \: x = \dfrac{ \sqrt{3} + 1 - ( \sqrt{3} - 1)}{2}$

$\\ \\ \sf \implies x = \dfrac{2 \sqrt{3} }{2} \: \: or \: x = \dfrac{ \sqrt{3} + 1 - \sqrt{3} + 1 }{2} \\ \\ \\ \sf \implies x = \sqrt{3} \: \: or \: \: x = \dfrac{2}{2} \\ \\ \\ \sf \implies x= \sqrt{3} \: \: or \: \: x = 1$

Therefore the value of x is √3 or 1.