Question

ax^2 +4x-a=0 solve by completing square method?​

Answer 1

ANSWER:

$a {x}^{2} + 4x - a = 0$

$a {x}^{2} + 4x = a$

$dividing \: by \: \: a \: \: on \: both \: sides$

$\frac{ax^{2} }{a} + \frac{4x}{a} = \frac{a}{a}$

${x}^{2} + \frac{4x}{a} = 1$

${(x)}^{2} + 2(x)( \frac{2}{a} ) = 1$

$adding \: {( \frac{2}{a} )}^{2} on \: both \: sides$

${(x)}^{2} + 2(x)( \frac{2}{a} ) + (\frac{ 2 }{a} )^{2} = 1 + ( { \frac{2}{a} })^{2}$

$using \: formula \: {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

${(x + \frac{2}{a}) }^{2} = 1 + \frac{4}{ {a}^{2} }$

${(x + \frac{2}{a}) }^{2} = \frac{ {a}^{2} }{ {a}^{2} } + \frac{4}{ {a}^{2} }$

${(x + \frac{2}{a}) }^{2} = \frac{ {a}^{2} + 4 }{ {a}^{2} }$

$taking \: squre \: root \: on \: both \: sides$

$\sqrt{ {(x + \frac{2}{a}) }^{2} } = \sqrt{ \frac{ {a}^{2} + 4 }{ {a}^{2} } }$

$x + \frac{2}{a} = \frac{ + }{} \frac{ \sqrt{ {a}^{2} + 4} }{ \sqrt{ {a}^{2} } }$

$x = - \frac{2}{a} \frac{ + }{} \frac{ \sqrt{ {a}^{2} + 4} }{a}$

$x = \frac{ - 2 \frac{ + }{} \sqrt{ {a}^{2} + 4} }{a}$

I hope it helps you. Please mark me as BRAINLIEST if you like.