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

Question

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

Sapling Academy · Accepted Answer

$4x^{2}-4a^{2}x+(a^{4}-b^{4})=0$ The Constant term $a^{4}-b^{4}$ can also be presented as: $(a^{2}-b^{2})(a^{2}+b^{2})$ Similarly, Coefficient of the middle term can be presented as: $-4a^{2}=-(2(a^{2}+b^{2})+2(a^{2}-b^{2}))$ Now, the original quadratic equation: $4x^{2}-4a^{2}x+(a^{4}-b^{4})=0$ $4x^{2}-(2(a^{2}+b^{2})+2(a^{2}-b^{2}))x+(a^{2}-b^{2})(a^{2}+b^{2})=0$ $4x^{2}-2(a^{2}+b{2})x-2(a^{2}-b^{2})x+(a^{2}-b^{2})(a^{2}+b^{2})=0$ $\{4x^{2}-2(a^{2}+b^{2})x\}-\{2(a^{2}-b^{2})x-(a^{2}-b^{2})(a^{2}+b^{2})\}=0$ $2x\{2x-(a^{2}+b^{2})\}-(a^{2}-b^{2})\{2x-(a^{2}+b^{2})\}=0$ $\{2x-(a^{2}+b^{2})\}\{2x-(a^{2}-b^{2})\}=0$ $2x-(a^{2}+b^{2})=0$  and  $2x-(a^{2}-b^{2})=0$ $\Rightarrow x=\frac{a^{2}+b^{2}}{2}$  and $x = \frac {a^2 - b^2}{2}$ Please do press "Heart" button if you liked the solution.

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

📚 Get Free Worksheet & Share Solution

Leave a Comment Cancel Reply

Contact us

Join us

Useful Link

Our Quizzes

📚 Get Free Worksheet & Share Solution

Related Math Questions

Leave a Comment Cancel Reply