Question

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

Answer 1

To solve this question we need to have the knowledge of quadratic equations. In this question we need to apply formulas for factoring the expression. The formula used to solve this will be${{x}^{2}}-{{y}^{2}}=\left( x-y \right)\left( x+y \right)$ . We will also take common values and factorize the expression accordingly.

Complete step-by-step solution:
The question ask us to find the value of $x$ in the expression $4{{x}^{2}}-4{{a}^{2}}x+{{a}^{4}}-{{b}^{4}}=0$using factorization method. We will expand the middle term in such a way that we get the common values for the calculation. The middle term will be expanded as:
$\Rightarrow 4{{x}^{2}}-2[({{a}^{2}}+{{b}^{2}})+({{a}^{2}}-{{b}^{2}})]x+({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})=0$
Since we know that the product of positive and negative is negative while that of negative and negative is positive and positive and positive is positive. Applying the same in the above equation we get:
$\Rightarrow 4{{x}^{2}}-2({{a}^{2}}+{{b}^{2}})x-2({{a}^{2}}-{{b}^{2}})x+({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})=0$
We will take common $2x$ and will factorize further:

$\Rightarrow 2x(2x-({{a}^{2}}+{{b}^{2}}))-({{a}^{2}}-{{b}^{2}})(2x-({{a}^{2}}+{{b}^{2}}))=0$

From the above term we can take $(2x-({{a}^{2}}+{{b}^{2}}))$ common. On doing this we get:

$\Rightarrow (2x-({{a}^{2}}+{{b}^{2}}))(2x-({{a}^{2}}-{{b}^{2}}))=0$
In the second term in the above expression we will take $4$ as common, hence resulting to:
$\Rightarrow (2x-({{a}^{2}}+{{b}^{2}}))(2x-({{a}^{2}}-{{b}^{2}}))=0$
We will now individually find the value for $x$ from the above expression.
$\Rightarrow (2x-({{a}^{2}}+{{b}^{2}}))=0$
$\Rightarrow x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2}$
The value of $x$for the another term will be:
$\Rightarrow (2x-({{a}^{2}}-{{b}^{2}}))=0$
$\Rightarrow x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$
So the two values of $x$ are $\dfrac{{{a}^{2}}+{{b}^{2}}}{2}$ and $\dfrac{{{a}^{2}}-{{b}^{2}}}{2}.$
$\therefore$ The values for $x$ for the following quadratic equation by factorization method: $4{{x}^{2}}-4{{a}^{2}}x+{{a}^{4}}-{{b}^{4}}=0$ are $\dfrac{{{a}^{2}}+{{b}^{2}}}{2}$ and $\dfrac{{{a}^{2}}-{{b}^{2}}}{2}.$

Note: We can check whether the answer we got is correct or not. To check this we will put the value of $x$ which are ${{a}^{2}}+{{b}^{2}}$ and in the equation $(x-(\dfrac{{{a}^{2}}+{{b}^{2}}}{2}))(x-(\dfrac{{{a}^{2}}-{{b}^{2}}}{2}))$. If the equation results in zero this means the answer is correct. On checking we get:
$\Rightarrow (x-(\dfrac{{{a}^{2}}+{{b}^{2}}}{2}))(x-(\dfrac{{{a}^{2}}-{{b}^{2}}}{2}))$
Putting $\dfrac{{{a}^{2}}+{{b}^{2}}}{2}$in place of $x$:
$\Rightarrow \left( \dfrac{{{a}^{2}}+{{b}^{2}}}{2}-\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{2} \right) \right)\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{2}-\left( \dfrac{{{a}^{2}}-{{b}^{2}}}{2} \right) \right)$
$\Rightarrow \left( \dfrac{{{a}^{2}}+{{b}^{2}}-{{a}^{2}}-{{b}^{2}}}{2} \right)\left( \dfrac{{{a}^{2}}+{{b}^{2}}-{{a}^{2}}+{{b}^{2}}}{2} \right)$
$\Rightarrow 0({{b}^{2}})=0$
Similarly on putting $\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$in the equation we get zero. So the answer is correct.