Question

find the roots of the quadratic equation by the method of completing the square: 4x2+x-3=0​

Answer 1

Answer:

4X^2 + 4x -3x -3 =0

4x(x-1) - 3(x-1)=0

(x-1)(4x-3)=0

x=1 ,3/4

Answer 2

$\large {\color{red}{ \bold{4 {x}^{2} + x - 3 = 0 }}}$

${\color{blue}{ \bold{ \implies \: 4 {x}^{2} + x =3}}}$

${\color{blue}{ \bold{ \implies \: {(2x)}^{2} + 2.2x. \frac{1}{4} + {( \frac{1}{4} )}^{2} - {( \frac{1}{4} )}^{2} =3}}}$

${\color{blue}{ \bold{ \implies {(2x + \frac{1}{4} )}^{2} - {( \frac{1}{16} )}=3}}}$

${\color{blue}{ \bold{ \implies {(2x + \frac{1}{4} )}^{2}=3 + { \frac{1}{16} }}}}$

${\color{blue}{ \bold{ \implies {(2x + \frac{1}{4} )}^{2}= { \frac{48 + 1}{16} }}}}$

${\color{blue}{ \bold{ \implies {(2x + \frac{1}{4} )}^{2}= { \frac{49}{16} }}}}$

${\color{blue}{ \bold{ \implies \sqrt{{(2x + \frac{1}{4} )}^{2}}= \ \: \sqrt{ \frac{49}{16} }}}}$

${\color{blue}{ \bold{ \implies \sqrt{{(2x + \frac{1}{4} )}^{2}}= \sqrt{ \frac{ {7}^{2} }{{4}^{2} } }}}}$

${\color{blue}{ \bold{ \implies {2x + \frac{1}{4} }= { \frac{ {7} }{{4} } }}}} \: \: \: \: \: \: \:or \: \: \: {\color{blue}{ \bold{ {2x + \frac{1}{4} }= - { \frac{ {7} }{{4} } }}}}$

${\color{blue}{ \bold{ \implies {2x }= { \frac{ {7} }{{4} } - \frac{1}{4} }}}} \: \: \: \: \: \: \:or \: \: \: {\color{blue}{ \bold{ {2x}= - { \frac{ {7} }{{4} } - \frac{1}{4} }}}} \:$

${\color{blue}{ \bold{ \implies {2x }= { \frac{ {6} }{{4} } }}}} \: \: \: \: \: \: \:or \: \: \: {\color{blue}{ \bold{ {2x}= { \frac{ { - 8} }{{4} } }}}}$

${\color{blue}{ \bold{ \implies {x }= { \frac{ {6} }{{8} } }}}} \: \: \: \: \: \: \:or \: \: \: {\color{blue}{ \bold{ {x}= { \frac{ { - 8} }{{8} } }}}}$

${\color{blue}{ \bold{ \implies {x }= { \frac{ {3} }{{4} } }}}} \: \: \: \: \: \: \:or \: \: \: {\color{blue}{ \bold{ {x}= { - 1 }}}}$

$\large \boxed{ \color{orange} \tt \bold{x = \frac{3}{4} \: \: \: and \: \: \: - 1}}$