Question

Solve the following quadratic equations by factorization:
abx²+(b²-ac)x-bc=0

Answer 1

The solved quadratic equation by Factorisation method is $abx^2+(b^2-4ac)x-bc=(bx-c)(ax+b)=0$

Step-by-step explanation:

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

To solve the given quadratic equation by factorisation method :

$abx^2+(b^2-ac)x-bc=0$

• Applying the Distributive property (x-y)z=xz-yz here $x=b^2$ ,y=ac and z=x
• $abx^2+(b^2)(x)-(ac)x-bc=0$
• $abx^2+b^2x-acx-bc=0$
• $bx(ax+b)-c(ax+b)=0$ ( by taking the common factor ) .

• $(bx-c)(ax+b)=0$ ( by combining the factors ) .

Therefore  $abx^2+(b^2-4ac)x-bc=(bx-c)(ax+b)=0$

The solved given quadratic equation by using Factorisation method is (bx-c)(ax+b) .