Question

Solve the equation 2x² + x - 4 = 0 by the method of competing square​

Answer 1

Answer

Given that,

$\sf {2x}^{2} + x - 4 = 0$

Multiplying 2 on both sides,

$\sf \dashrightarrow {4x}^{2} + 2x - 8 = 0$

Shifting 8 from LHS to RHS,

$\sf \dashrightarrow {4x}^{2} + 2x = 8$

$\sf \dashrightarrow {(2x)}^{2} + 2 \times 2x \times \dfrac{1}{2} + \bigg( \dfrac{1}{2} \bigg)^{2} = 8 + \bigg( \dfrac{1}{2} \bigg)^{2}$

Add $\sf \bigg( \dfrac{1}{2} \bigg)^2$ on both sides.

$\sf \dashrightarrow \bigg( 2x + \dfrac{1}{2} \bigg)^2 = \bigg( 8 + \dfrac{1}{4} \bigg) = \dfrac{33}{4} = \bigg( \dfrac{\sqrt{33}}{2} \bigg)^{2}$

Take the square root on both sides.

$\sf \dashrightarrow 2x + \dfrac{1}{2} = \pm \bigg( \dfrac{\sqrt{33}}{2} \bigg)$

$\sf \dashrightarrow 2x + \dfrac{1}{2} = \dfrac{\sqrt{33}}{2} \: or \: 2x + \dfrac{1}{2} = \dfrac{- \sqrt{33}}{2}$

$\sf \dashrightarrow 2x = \bigg( \dfrac{\sqrt{33}}{2} - \dfrac{1}{2} \bigg) = \dfrac{( \sqrt{33} - 1)}{2}$

$\sf \dashrightarrow 2x = \dfrac{- \sqrt{33}}{2} - \dfrac{1}{2} = \dfrac{- ( \sqrt{33} + 1)}{2}$

Shifting 2 from LHS to RHS,

$\sf \dashrightarrow x = \dfrac{( \sqrt{33} - 1)}{4} \: or \: x = \dfrac{- ( \sqrt{33} + 2)}{4}$

Thus, $\sf \dfrac{-1 + \sqrt{33}}{4}$ and $\sf \dfrac{- 1 - \sqrt{33}}{4}$ are the roots of the given equation.