Q)  Solve the following quadratic equation by factorization method: 4x^2-2(a^2+b^2)x+a^2b^2=0



The Constant term a^2b^2 can also be presented as:

a^2\times b^2

Similarly, Coefficient of the middle term can be presented as:

-2 (a^2 + b^2)

Now, the original quadratic equation:


\implies 4x^{2}-2a^{2}x-2b^{2}x+a^{2}b^{2}=0

\implies (4x^{2}-2a^{2}x)-(2b^{2}x-a^{2}b^{2})=0

\implies 2x(2x-a^{2})-b^{2}(2x-a^{2})=0

\implies (2x-a^{2})(2x-b^{2})=0

\implies (2x-a^{2})=0


\implies  x=\frac{a^{2}}{2} ~or~ x=\frac{b^{2}}{2}

Please do press “Heart” button if you liked the solution. 

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top