Question

HEY❤
BRAINLIEST ANSWER

solve the following by factorization...
(a+b)² x² - 4abx - (a-b)² = 0

sol... x = 1 , x = -(a-b)/(a+b)...

acc. to RD Sharma (2013)​

Answer 1

Step-by-step explanation:

$Method\:1\huge\mathcal{Given}\\\mathcal{(a+b)^2x^2-4abx-(a-b)^2=0\rightarrow{Eq:01}}\\By\:observation\:we\:can\:write\:4ab\:as(a+b)^2-(a-b)^2\\(a+b)^2-(a-b)^2=4ab\\(a-b)^2-(a+b)^2=-4ab\\By\:this\:Equation\:01\:becomes\:\\(a+b)^2x^2+(a-b)^2x-(a+b)^2x-(a-b)^2=0\\(a+b)^2x[x-1]+(a-b)^2[x-1]=0\\(x-1)[(a+b)^2x+(a-b)^2]=0\\\huge{x-1=0=>x=1}\\(a+b)^2x+(a-b)^2=0\\\huge{x=-\left(\frac{a-b}{a+b}\right)^2} \\ \\ Method\:2\huge\mathcal{Given}\\(a+b)^2x^2-4abx-(a-b)^2=0\\x=\frac{-(-4ab)\pm\sqrt{(-4ab)^2-4.(a+b)^2.(-(a-b)^2)}}{2.(a+b)^2}\\x=\frac{4ab\pm\sqrt{16a^2b^2+4(a^2-b^2)^2}}{2(a+b)^2}\\x=\frac{4ab\pm\sqrt{16a^2b^2+4a^4-8a^2b^2+4b^4}}{2(a+b)^2}\\x=\frac{4ab\pm\sqrt{4(a^4+2a^2b^2+b^4)}}{2(a+b)^2}\\x=\frac{4ab\pm\sqrt{4(a^2+b^2)^2}}{2(a+b)^2}\\x=\frac{4ab\pm2(a^2+b^2)}{2(a+b)^2}\\Case:01\\x=\frac{4ab+2a^2+2b^2}{2(a+b)^2}\\x=\frac{2(a+b)^2}{2(a+b)^2}\\\huge\mathcal{x=1}\\Case:02\\x=\frac{4ab-2a^2-2b^2}{2(a+b)^2}\\x=\frac{-2(a^2-2ab+b^2)}{2(a+b)^2}\\x=\frac{-2(a-b)^2}{2(a+b)^2}\\x=\frac{-(a-b)^2}{(a+b)^2}\\\huge\mathcal{x=-\left(\frac{a-b}{a+b}\right)^2}$

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Answer 2

Answer:

HEY MATE HERE IS UR ANSWER

(a+b)² x² - 4abx - (a-b)² = 0

(a+b)² x² - (a+b)²x +(a-b)²x - (a-b)² = 0

as a^2+b^2+2ab-a^2-b^2+2ab

=4ab

so,

(a+b)² x(x-1)+ (a-b)² (x-1)=0

(x (a+b)² +(a-b)² )(x-1)=0

hence,

x have two solution;-);-)

x= -(a-b)² / (a+b)²

and

x=1