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

Hint: 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.