Question

Solve the quadratic $4{x^2} + 4bx - \left( {{a^2} - {b^2}} \right) = 0$ for the value of x.

Answer 1

Hint: Here, we will proceed by rearranging the terms in the LHS of the given quadratic equation so that we can use the formula ${\left( {y + z} \right)^2} = {y^2} + 2yz + {z^2}$. From there we will get two linear equations in variable x which can be easily solved for the values of x.

Complete step-by-step answer:

Given quadratic equation is $4{x^2} + 4bx - \left( {{a^2} - {b^2}} \right) = 0{\text{ }} \to {\text{(1)}}$
$ \Rightarrow 4{x^2} + 4bx - {a^2} + {b^2} = 0$
Taking the term $ - {a^2}$ from the LHS to the RHS of the above equation, we get
$ \Rightarrow 4{x^2} + 4bx + {b^2} = {a^2}{\text{ }} \to {\text{(2)}}$
In the above equation, the term $4{x^2}$ can also be written as ${\left( {2x} \right)^2}$ and the term $4bx$ can also be written as ${\text{2}}\left( {2x} \right)\left( b \right)$ so finally equation (2) can be written as
$ \Rightarrow {\left( {2x} \right)^2} + {\text{2}}\left( {2x} \right)\left( b \right) + {b^2} = {a^2}$
Using the formula ${\left( {y + z} \right)^2} = {y^2} + 2yz + {z^2}$ in the above equation, we get
$
   \Rightarrow {\left( {2x + b} \right)^2} = {a^2} \\
   \Rightarrow \left( {2x + b} \right) = \pm \sqrt {{a^2}} \\
   \Rightarrow 2x + b = \pm a \\
 $
Either $
  2x + b = a \\
   \Rightarrow 2x = a - b \\
   \Rightarrow x = \dfrac{{a - b}}{2} \\
 $ or $
  2x + b = - a \\
   \Rightarrow 2x = - a - b \\
   \Rightarrow x = - \left( {\dfrac{{a + b}}{2}} \right) \\
 $
Therefore, the roots of the given quadratic equation are $x = \dfrac{{a - b}}{2}$ and $x = - \left( {\dfrac{{a + b}}{2}} \right)$.

Note: In this particular problem, the roots of the given quadratic equation i.e., $4{x^2} + 4bx - \left( {{a^2} - {b^2}} \right) = 0$ can also be solved by comparing this equation with the general form of any quadratic equation i.e., $a{x^2} + bx + c = 0$ and then using the discriminant method i.e., $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.