Question

On solving the given equation $4{{x}^{2}}-4{{a}^{2}}x+\left( {{a}^{4}}-{{b}^{4}} \right)=0$ we get the value of x is equal to
$\left( a \right)\dfrac{{{a}^{2}}\pm {{b}^{2}}}{2}$
$\left( b \right)\dfrac{{{a}^{2}}}{2}$
$\left( c \right)\dfrac{{{b}^{2}}}{2}$
$\left( d \right)\dfrac{{{a}^{2}}\div {{b}^{2}}}{2}$

Answer 1

To solve this question we will use the standard quadratic equation and formula of calculating its root. The equation is $A{{x}^{2}}+Bx+C,$discriminant $D={{B}^{2}}-4AC$ and the value of x, $x=\dfrac{-B\pm \sqrt{D}}{2A}.$ Substituting all the values in these, we will get the result.

Complete step-by-step solution
We are given the equation as
$4{{x}^{2}}-4{{a}^{2}}x+\left( {{a}^{4}}-{{b}^{4}} \right)=0......\left( i \right)$
Now, this equation is quadratic as it has a greater power of x as 2. So, we will compare the given equation (i) with the standard quadratic equation. The standard quadratic equation is given by
$A{{x}^{2}}+Bx+C=0......\left( ii \right)$
Comparing the equations (i) and (ii), we observe that
$A=4$
$B=-4{{a}^{2}}$
$C={{a}^{4}}-{{b}^{4}}$
The discriminant D of the equation is given by the formula $D={{B}^{2}}-4AC$ where D is called the discriminant of the quadratic equation. Now, calculating the discriminant D of equation (i), we have,
$D={{\left( -4a \right)}^{2}}-4\times 4\times \left( {{a}^{4}}-{{b}^{4}} \right)$
We know that the square of a negative integer is always positive. Now, taking the square of 4 as 16, we get
$\Rightarrow D=16{{a}^{4}}-16{{a}^{4}}+16{{b}^{4}}$
We can cancel the similar terms, then we will get
$\Rightarrow D=16{{b}^{4}}$
Now, finally calculating x using the formula
$x=\dfrac{-B\pm \sqrt{D}}{2A}$
Substituting the value of D, B and A, we have,
$\Rightarrow x=\dfrac{-\left( -4{{a}^{2}} \right)\pm \sqrt{16{{b}^{4}}}}{2\times 4}$
$\Rightarrow x=\dfrac{4{{a}^{2}}\pm 4{{b}^{2}}}{8}$
Taking 4 common, we have,
$\Rightarrow x=\dfrac{{{a}^{2}}\pm {{b}^{2}}}{2}$
So, the value of x is:
$\Rightarrow x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2};x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$
Hence, option (a) is the right answer.

Note: Because, the quadratic equation has degree 2, means it always has 2 roots. There is a possibility that the two roots are equal. But always there are 2 roots. So, the possibilities of the options (b) $\dfrac{{{a}^{2}}}{2}$ and (c) $\dfrac{{{b}^{2}}}{2}$ and (d) $\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$ are eliminated.