Question

Solve the following quadratic equation using the quadratic formula: ${a^2}{x^2} + ({a^2} - {b^2})x - {b^2} = 0$
$\left( a \right){\text{ }}\left\{ { - 1,\dfrac{{{b^2}}}{{{a^2}}}} \right\}$
$\left( b \right){\text{ }}\left\{ {1,\dfrac{{{b^2}}}{{{a^2}}}} \right\}$
$\left( c \right){\text{ }}\left\{ { - 1,\dfrac{{ - {b^2}}}{{{a^2}}}} \right\}$
\[\left( d \right){\text{ None of these}}\]

Answer 1

Hint:
For solving this first of all we find the values $a, b, c$ by comparing the equation with the general quadratic equation which is $a{x^2} + bx + c = 0$ and then by using the values we will put the values in $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and on solving we will get the values of $x$.

Formula used:
As we know for the quadratic equation$a{x^2} + bx + c = 0$,
The quadratic formula will be given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
we have the equation given in the question as ${a^2}{x^2} + ({a^2} - {b^2})x - {b^2} = 0$
Now on comparing the above equation with the general quadratic equation, we get
$a = {a^2},{\text{ b = }}\left( {{a^2} - {b^2}} \right),{\text{ c = - }}{{\text{b}}^2}$
Therefore, on substituting the above values in$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - \left( {{a^2} - {b^2}} \right) \pm \sqrt {{{\left( {{a^2} - {b^2}} \right)}^2} - 4{a^2}\left( { - {b^2}} \right)} }}{{2{a^2}}}$
Now on solving the above equation, we get
$ \Rightarrow x = \dfrac{{ - \left( {{a^2} - {b^2}} \right) \pm \sqrt {\left( {{a^4} - {b^4} - 4{a^2}{b^2}} \right) + 4{a^2}{b^2}} }}{{2{a^2}}}$
So it can also be written as
\[ \Rightarrow x = \dfrac{{ - \left( {{a^2} - {b^2}} \right) \pm \sqrt {{{\left( {{a^2}} \right)}^2} + {{\left( {{b^2}} \right)}^2}} }}{{2{a^2}}}\]
And also on removing the square root, we get
\[ \Rightarrow x = \dfrac{{ - \left( {{a^2} - {b^2}} \right) \pm \left( {{a^2}} \right) + \left( {{b^2}} \right)}}{{2{a^2}}}\]
So now we will solve for both sign separately, we get
$ \Rightarrow x = \dfrac{{ - \left( {{a^2} - {b^2}} \right) + \left( {{a^2}} \right) + \left( {{b^2}} \right)}}{{2{a^2}}}{\text{ or }}x = \dfrac{{ - \left( {{a^2} - {b^2}} \right) - \left( {{a^2}} \right) - \left( {{b^2}} \right)}}{{2{a^2}}}$
Now on expanding the above equation, we get
$ \Rightarrow x = \dfrac{{ - {a^2} + {b^2} + \left( {{a^2}} \right) + \left( {{b^2}} \right)}}{{2{a^2}}}{\text{ or }}x = \dfrac{{ - {a^2} + {b^2} - \left( {{a^2}} \right) - \left( {{b^2}} \right)}}{{2{a^2}}}$
Therefore, on solving the above equation and also since the same term having opposite sign will cancel each other, so we get
$ \Rightarrow x = \dfrac{{2{b^2}}}{{2{a^2}}}{\text{ or }}x = \dfrac{{ - 2{a^2}}}{{2{a^2}}}$
Now on solving the above equation, we get
$ \Rightarrow x = \dfrac{{{b^2}}}{{{a^2}}}{\text{ or }}x = - 1$
Therefore, the values $x$will be equal to $ - 1,\dfrac{{{b^2}}}{{{a^2}}}$

Hence, the option $\left( a \right)$ is correct.

Additional information:
One thing though, you do not need both $a,b,c$ to be different from zero, all you need is $a$ to be different from zero.$a$ Different from zero ensures that you have a quadratic equation, but you can have either $b$ or $c$ both equal to zero.

Note:
This type of question requires a lot of calculation so while doing the calculation we should be aware and also we have to little bit memorize the formula for finding the values as we had used some formulas which we have to keep in mind while solving it.