Question

Solve the following quadratic equations by factorization:
(a+b)²x²-4abx-(a-b)²=0

Answer 1

The solved given quadratic equation is $(a+b)^2x^2-4abx-(a-b)^2=(a^2+b^2-2ab)(x-1)$

Step-by-step explanation:

Given quadratic equation is $(a+b)^2x^2-4abx-(a-b)^2=0$

To solve the given quadratic equation by Factorization method :

$(a+b)^2x^2-4abx-(a-b)^2=0$

Expanding the powers by using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$  

$(a^2+2ab+b^2)x^2-4abx-(a^2-2ab+b^2)=0$

Using the Distributive property (x+y+z)a=xa+ya+za

• $a^2(x^2)+2ab(x^2)+b^2x^2-4abx-a^2-(-2ab)-(b^2)=0$

• $a^2x^2+2abx^2+b^2x^2-4abx-a^2+2ab-b^2=0$
• $a^2(x-1)+b^2(x-1)-2abx+2ab=0$ ( taking the common terms outside the factors )

• $(a^2+b^2)(x-1)-2ab(x-1)=0$ ( taking the common factors outside the terms )
• $(a^2+b^2-2ab)(x-1)=0$ ( taking the common factors outside the terms )

Therefore $(a+b)^2x^2-4abx-(a-b)^2=(a^2+b^2-2ab)(x-1)=0$

Therefore the solved given quadratic equation is $(a^2+b^2-2ab)(x-1)$

Answer 2

hope you understand.............