Question

squaring 2x²-5x +3 =0 by complete squaring method​

Answer 1

$2 {x}^{2} - 5x + 3 = 0$

Multiply both sides of equation by 2 .

$4 {x}^{2} - 10x + 6 = 0 \\ = > {(2x)}^{2} - 2 \times 2x \times \frac{5}{2} + { (\frac{5}{2}) }^{2} - { (\frac{5}{2}) }^{2} + 6 = 0 \\ = > {(2x - \frac{5}{2}) }^{2} - \frac{1}{4} = 0 \\ = > {(2x - \frac{5}{2}) }^{2} = \frac{1}{4} \\ = > (2x - \frac{5}{2}) = \frac{1}{2} \\ = > 2x = \frac{1}{2} - \frac{5}{2} \\ = > x = \frac{ - 2}{2}$

x = -1

Answer 2

Correct question :-

Factorise 2x² - 5x + 3 = 0 by completing the square method.

Solution :-

$2x^2\:-5x\:+\:3\:=\:0 \\ \\ Dividing\:both\:side\:by\:2 \\ \\ => \: x^2\:-\: \dfrac{5x}{2}\:+\: \dfrac{3}{2}\:=\:0 \\ \\ \\ x^2\:-\:2 \times \dfrac{5x}{4}\:= \: - \dfrac{3}{2}\:$

$Adding\:{(\dfrac{5}{4})}^2 \:to\:both\:sides \\ \\ \\ => \:x^2\:-\:2\times \dfrac{5x}{4}\: + \: {(\dfrac{5}{4})}^2 \: = :\- \dfrac{3}{2}\:+\:{(\dfrac{5}{4})}^2 \\ \\ \\ {(x\:-\: \dfrac{5}{4})}^2\:=\: \dfrac{1}{16}$

$=> {(x\:-\: \dfrac{5}{4})}^2\:=\: {( \dfrac{1}{4})}^2 \\ \\ => \: x\:- \dfrac{5}{4}\:=\: \dfrac{1}{4}\:or\: - \dfrac{1}{4} \\ \\ \\ => \: x\:=\: \dfrac{3}{2}\:or\:1$

$\boxed{\green{Hence,\:x\:=\: \dfrac{3}{2}\:or\:x\:=\:1}}$