Math, asked by Aartikmari230, 1 year ago

Find the roots of the quadratic equation 4x square + 4bx - (a square - b square = 0 by the metod of completing the square .

Answers

Answered by clearbandit
0
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Answered by VelvetBlush
6

\bigstar{\huge{\underline{\mathsf{\red{Answer}}}}}

Given :-  \sf{{4x}^{2}  + 4bx - ( {a}^{2}  -  {b}^{2}) = 0 }

\longrightarrow\sf{ {x}^{2}  + bx - ( \frac{ {a}^{2} -  {b}^{2} ) }{4}  = 0}

\longrightarrow\sf{ {x}^{2}  + bx =  \frac{ {a}^{2} -  {b}^{2}  }{4}}

\longrightarrow \sf{{x}^{2}  + 2( \frac{b}{2} )x +   {( \frac{b}{2}) }^{2}   =  \frac{ {a}^{2}  -  {b}^{2} }{4}  +  { (\frac{b}{2} )}^{2} }

\longrightarrow\sf{ {(x +  \frac{b}{2}) }^{2}  =  \frac{ {a}^{2} -  {b}^{2}   +  {b}^{2} }{4}}

\longrightarrow\sf{ { (x + \frac{b}{2}) }^{2}  =  \frac{ {a}^{2} }{4}  = x +  \frac{b}{2}  = ± \frac{a}{2}  = x =±  \frac{a}{2}  -  \frac{b}{2}}

\longrightarrow\sf{x =  \frac{a - b}{2}  \:or \: x =  -  (\frac{a + b}{2} )}

Hence, the roots of the given equation are

\sf{( \frac{a - b}{2} ) \: and \:  - ( \frac{a + b}{2} )}

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