Question

The roots of $ {x^2} + 2ax + \left( {{a^2} - {b^2}} \right) = 0 $ are

Answer 1

Hint: In this question, we have to find the roots of the quadratic equation $ {x^2} + 2ax + \left( {{a^2} - {b^2}} \right) = 0 $ . As it is a quadratic equation, it will have two roots. Let these two roots be $ \alpha $ and $ \beta $ . Now, a quadratic equation can be solved using the quadratic formula. The quadratic formula is
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Where, $ a = 1 $ , $ b = 2a $ and $ c = \left( {{a^2} - {b^2}} \right) $
Substituting these values in quadratic formula, we will get the values of $ \alpha $ and $ \beta $ .

Complete step-by-step answer:
In this question, we are given a quadratic equation, and we need to find its roots.
The given quadratic equation is: $ {x^2} + 2ax + \left( {{a^2} - {b^2}} \right) = 0 $
Now, a quadratic equation always has 2 roots. Hence, our given equation will have two roots. Let these two roots be $ \alpha $ and $ \beta $ .
Now, to solve this quadratic equation, we are going to use the quadratic formula. The quadratic formula is
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] - - - - - - - - - - - - (1)
Here, $ a = 1 $ , $ b = 2a $ and $ c = \left( {{a^2} - {b^2}} \right) $ . Therefore, substituting these values in equation (1), we get
\[
   \Rightarrow x = \dfrac{{ - 2a \pm \sqrt {{{\left( {2a} \right)}^2} - 4\left( 1 \right)\left( {{a^2} - {b^2}} \right)} }}{{2\left( 1 \right)}} \\
   \Rightarrow x = \dfrac{{ - 2a \pm \sqrt {4{a^2} - 4\left( {{a^2} - {b^2}} \right)} }}{2} \\
   \Rightarrow x = \dfrac{{ - 2a \pm \sqrt {4{a^2} - 4{a^2} + 4{b^2}} }}{2} \\
   \Rightarrow x = \dfrac{{ - 2a \pm \sqrt {4{b^2}} }}{2} \\
   \Rightarrow x = \dfrac{{ - 2a \pm 2b}}{2} \\
 \]
Therefore, $ \alpha = \dfrac{{ - 2a + 2b}}{2} $ and $ \beta = \dfrac{{ - 2a - 2b}}{2} $
 $
   \Rightarrow \alpha = \dfrac{{ - 2a + 2b}}{2} \\
   \Rightarrow \alpha = \dfrac{{2\left( {b - a} \right)}}{2} \\
   \Rightarrow \alpha = \left( {b - a} \right) \;
  $
And
 $
   \Rightarrow \beta = \dfrac{{ - 2a - 2b}}{2} \\
   \Rightarrow \beta = \dfrac{{ - 2\left( {a + b} \right)}}{2} \\
   \Rightarrow \beta = - \left( {a + b} \right) \;
  $
Hence, the two roots of the given quadratic equation are $ \left( {b - a} \right) $ and $ - \left( {a + b} \right) $ .
So, the correct answer is “ $ \left( {b - a} \right) $ and $ - \left( {a + b} \right) $ .”.

Note: Here, $ \sqrt {{b^2} - 4ac} $ is known as the discriminant of a quadratic equation. Nature of roots can be found based on the value of discriminant of a quadratic equation.
If $ D > 1 $ , then the roots are real and distinct.
If $ D = 0 $ , then the roots of the equation are 0.
If $ D < 0 $ , then the roots of the equation are imaginary.