Question

A plus b the whole square x square + 8 a square minus b square into X + 16 a minus b the whole square equal to zero solve for x

Answer 1

$\text{I have solved this quadratic equation by factorization method by splitting its middle term}$

$\text{Given equation is}$

$\bf(a+b)^2x^2+8(a^2-b^2)x+16(a-b)^2=0$

$\implies(a+b)^2x^2+4(a^2-b^2)x+4(a^2-b^2)x+16(a-b)^2=0$

$\implies(a+b)^2x^2+4(a+b)(a-b)x+4(a-b)(a+b)x+16(a-b)^2=0$

$\implies(a+b)x[(a+b)x+4(a-b)]+4(a-b)[(a+b)x+4(a-b)]=0$

$\implies[(a+b)x+4(a-b)][(a+b)x+4(a-b)]=0$

$\implies[(a+b)x+4(a-b)]=0$

$\implies\,x=\displaystyle\frac{-4(a-b)}{a+b}\;\text{(twice)}$

$\therefore\textbf{The solution set is \{\displaystyle\frac{-4(a-b)}{a+b}\} }$

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Answer 2

from links anwer

Step-by-step explanation:

a1ba5a089049eda7879f3c59ad6f684d.pdf