Question

Find the roots of the following equation $4{x^2} + 4bx - \left( {{a^2} - {b^2}} \right) = 0$ .

Answer 1

Hint: First, compare the given equation $4{x^2} + 4bx - \left( {{a^2} - {b^2}} \right) = 0$ with $a{x^2} + bx + c = 0$ and find a, b and c.
Now, find discriminant $D = \sqrt {{b^2} - 4ac} $ .
Thus, the roots of equation can be found using $\dfrac{{ - b + D}}{{2a}}$ and $\dfrac{{ - b - D}}{{2a}}$ .

Complete step-by-step answer:
The given equation is $4{x^2} + 4bx - \left( {{a^2} - {b^2}} \right) = 0$ and we have to find its roots.
To find roots, first, we have to find the discriminant D of the equation.
 $
  \therefore {D^2} = {b^2} - 4ac \\
  \therefore {D^2} = {\left( {4b} \right)^2} - 4\left( 4 \right)\left( { - \left( {{a^2} - {b^2}} \right)} \right) \\
  \therefore {D^2} = 16{b^2} + 16{a^2} - 16{b^2} \\
  \therefore {D^2} = 16{a^2} \\
 $
Now, taking square root on both sides
 $
  \therefore \sqrt {{D^2}} = \sqrt {16{a^2}} \\
  \therefore D = 4a \\
 $
We have \[D = 4a\] .
Now, the roots of equation are $\dfrac{{ - b + D}}{{2a}}$ and $\dfrac{{ - b - D}}{{2a}}$ .
 $\therefore \dfrac{{ - b + D}}{{2a}} = \dfrac{{ - 4b + 4a}}{{2\left( 4 \right)}}$
          $
   = \dfrac{{4\left( {a - b} \right)}}{{2\left( 4 \right)}} \\
   = \dfrac{{a - b}}{2} \\
 $
And
 $
  \dfrac{{ - b - D}}{{2a}} = \dfrac{{ - 4b - 4a}}{{2\left( 4 \right)}} \\
   = \dfrac{{ - 4\left( {a + b} \right)}}{{2\left( 4 \right)}} \\
   = \dfrac{{ - \left( {a + b} \right)}}{2} \\
 $
Thus, the roots of equation are $\dfrac{{a - b}}{2}$ and $\dfrac{{ - \left( {a + b} \right)}}{2}$.


Note: To verify whether the answer is correct or not, we can do as follows.
Product of roots of any equation $ = \dfrac{c}{a}$ .
So, here, the product of roots must be equal to $\dfrac{{ - \left( {{a^2} - {b^2}} \right)}}{4}$ .
 $\therefore \dfrac{{a - b}}{2} \times \dfrac{{ - \left( {a + b} \right)}}{2} = \dfrac{{ - \left( {{a^2} - {b^2}} \right)}}{4}$
Thus, the product of the roots of the given equation is equal to $\dfrac{{ - \left( {{a^2} - {b^2}} \right)}}{4}$ .
Now, sum of the roots of any equation $ = \dfrac{{ - b}}{a}$ .
So, here, the sum of roots of the equation must be equal to $\dfrac{{ - 4b}}{4} = - b$ .
 $\therefore \dfrac{{a - b}}{2} + \dfrac{{ - \left( {a + b} \right)}}{2} = \dfrac{{a + b - a - b}}{2} = \dfrac{{ - 2b}}{2} = - b$