Math, asked by Preru14, 1 year ago

Solve quadratic equation by Factorization:


4x {}^{2}  + 4bx - (a {}^{2}  - b {}^{2} ) = 0

Answers

Answered by Anonymous
4
Hey there !!


▶ The given quadratic equation :-

4x {}^{2} + 4bx - (a {}^{2} - b {}^{2} ) = 0

=> 4x² + 4bx = ( a² - b² ) .

[Adding b² on both side ]

=> 4x² + 4bx + b² = a² - b² + b² .

=> (2x)² + 2 × 2x × b + b² = a² .

=> ( 2x + b )² = a² .

[ Taking square root on both side, we get ] .

=> 2x + b = ± a.

=> 2x + b = a or 2x + b = -a .

=> 2x = a - b or 2x = - a - b .

  \boxed{ \bf \therefore x =  \frac{a - b}{2}  \:  \:  \: or \:  \:  \:  x =  -  \frac{a + b}{2} . }\\


✔✔ Hence, it is solved ✅✅.



THANKS



#BeBrainly.
Answered by Anonymous
0

\huge\mathbb{SOLUTION:-}

\mathtt{4x {}^{2}  + 4bx - \bigg(a {}^{2}  - b {}^{2} \bigg) = 0}

:\implies x {}^{2}  + bx - \bigg( \frac{a {}^{2}  - b {}^{2} }{4} \bigg) = 0

:\implies x {}^{2}  + 2\bigg( \frac{b}{2}\bigg) x =  \frac{a {}^{2}  - b {}^{2} }{4}

:\implies x {}^{2}  + 2\bigg(\frac{b}{2} \bigg)x + \bigg( \frac{b}{2} \bigg) {}^{2}  =  \frac{a {}^{2} - b {}^{2}  }{4}  + \bigg( \frac{b}{2}\bigg){}^{2}

:\implies \bigg(x +  \frac{b}{2} \bigg){}^{2} =  \frac{a {}^{2} }{4}

:\implies x +  \frac{b}{2}  =  \pm \frac{a}{2}

:\implies x =  \frac{ - b}{2}  \pm \frac{a}{2}

:\implies x =  \frac{ - b \:  - a}{2} , \frac{ - b + a}{2}

Hence, \: the \: roots \: are\implies \:  -\bigg( \frac{a + b}{2} \bigg) \: and \: \bigg( \frac{a - b}{2} \bigg)

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