Question

solve 2 x square - 5 x minus 3 equal to zero squaring method​

Answer 1

x = 3 , x = - ¹/₂

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$\orange{\bigstar}$  Given  $\green{\bigstar}$

2x² - 5x - 3 = 0

$\orange{\bigstar}$  To Find  $\green{\bigstar}$

To solve given equation by completing square method

$\orange{\bigstar}$  Formula Applied  $\green{\bigstar}$

$\pink{\bigstar}\ \; \bf (a-b)^2=a^2-2ab+b^2$

$\orange{\bigstar}$  Solution  $\green{\bigstar}$

$\bf 2x^2-5x-3=0$

Divide both sides of equation by 2 ,

$\to \rm \dfrac{2x^2-5x-3}{2}=\dfrac{0}{2}\\\\\to \rm \dfrac{2x^2}{2}-\dfrac{5x}{2}-\dfrac{3}{2}=0\\\\\to \rm x^2-2.x.\dfrac{5}{4}=\dfrac{3}{2}$

Compare 2.x.5/4 with 2ab , we will get , b = 5/4

So , Add (5/4)² on both sides ,

$\to \rm x^2-2.x.\dfrac{5}{4}+\left( \dfrac{5}{4}\right)^2=\dfrac{3}{2}+\left( \dfrac{5}{4}\right)^2$

LHS resemble's the formula , a² - 2ab + b² , so ,

$\to \rm \left( x-\dfrac{5}{4}\right)^2=\dfrac{3}{2}+\dfrac{25}{16}\\\\\to \rm \left(x-\dfrac{5}{4}\right)^2=\dfrac{24+25}{16}\\\\\to \rm \left(x-\dfrac{5}{4}\right)^2=\dfrac{49}{16} \\\\\to \rm \left(x-\dfrac{5}{4}\right)=\pm\sqrt{\dfrac{49}{16}}\\\\\to \rm \left(x-\dfrac{5}{4}\right)=\pm \dfrac{7}{4}\\\\\to \bf x=\dfrac{5\pm 7}{4}\ \; \pink{\bigstar}$

$\to \rm x=\dfrac{5+7}{4}\ \; \; \&\ \; \; x=\dfrac{5-7}{4}\\\\\to \rm x=\dfrac{12}{4}\ \; \; \&\ \; \; x=-\dfrac{2}{4}\\\\\to \rm x=3\ \; \; \&\ \; \; x=-\dfrac{1}{2}\ \; \pink{\bigstar}$

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Answer 2

• We have to find the roots by squaring method !

$\sf\ \ \ \ 2x^2-5x-3=0\\ \\ \\ \sf\ \ \ \small\ dividing \ By \ 2\\ \\ \\ \implies\sf\ \ \dfrac{2x^2-5x-3}{2}= \dfrac{0}{2}\\ \\ \\\implies\sf\ x^2-\dfrac{5}{2}x-\dfrac{3}{2}=0\\ \\ \\ \implies\sf\ x^2- 2\times \dfrac{5}{4}x= \dfrac{3}{2}\\ \\ \\\scriptsize\sf\ \ (a+b)^2= a^2+2ab+b^2\ \\ \\ \\ \implies\sf 2ab= 2\times \dfrac{5}{4}x\ \ \ ,\ b= \dfrac{5}{4}\\ \\ \\ \implies\sf adding \ (^5\!/_4)^2\ \ both\ side \\ \\ \\ \implies\sf x^2-2.\dfrac{5}{4}x+\Big[\dfrac{5}{4}\Big]^2=\dfrac{3}{2}+\Big[\dfrac{5}{4}\Big]^2\\ \\ \\ \implies\sf \bigg[x-\dfrac{5}{4}\bigg]^2= \dfrac{3}{2}+\dfrac{25}{16}\\ \\ \\ \implies\sf \bigg[x-\dfrac{5}{4}\bigg]^2= \dfrac{24+25}{16}\\ \\ \\ \implies\sf \bigg[x-\dfrac{5}{4}\bigg]= \pm \sqrt{\dfrac{49}{16}}\\ \\ \\ \implies\sf x-\dfrac{5}{4}= \pm\bigg[\dfrac{7}{4}\bigg]\\ \\ \\\implies\sf x= \dfrac{5\pm7}{4}\\ \\ \\ \implies\sf \ x= \dfrac{-2}{4}= \bigg[\dfrac{-1}{2}\bigg]\ \ ;\ \ or\ x= \bigg[\dfrac{12}{4}\bigg]= 3$

$\underline{\boxed{\sf\ \ x= ( ^{-1}\!/_2)\ \ or\ \ 3}}$