Question

If the quadratic equation $\left( {{b}^{2}}+{{c}^{2}} \right){{x}^{2}}-2\left( a+b \right)cx+\left( {{c}^{2}}+{{a}^{2}} \right)=0$ has equal roots. Then find the relation between a, b, c.

Answer 1

Hint: To solve this question, we will find the roots of a standard quadratic equation by one of the three methods available. We will then write an equation of equal roots and derive a condition for the same. Then, we will apply the condition in the given quadratic equation and try to find the relation between a, b and c.

Complete step-by-step solution:
A standard quadratic equation is in the form $a{{x}^{2}}+bx+c=0$.
There are three methods to find the roots of a quadratic equation, viz. factorization method, completing the square method, and formula method. All three methods will yield the same root and thus it doesn’t matter which method is used to solve the equation.
We will use formula method to solve the equation $a{{x}^{2}}+bx+c=0$.
According to the formula method, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Thus, the two roots are $x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ and $x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$
Now, if the roots are equal, then $\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$
We will split the numerator:
$\Rightarrow \dfrac{-b}{2a}+\dfrac{\sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b}{2a}-\dfrac{\sqrt{{{b}^{2}}-4ac}}{2a}$
$\dfrac{-b}{2a}$ cancels out from both sides.
\[\begin{align}
  & \Rightarrow \dfrac{2\sqrt{{{b}^{2}}-4ac}}{2a}=0 \\
 & \Rightarrow \sqrt{{{b}^{2}}-4ac}=0 \\
 & \Rightarrow {{b}^{2}}-4ac=0 \\
\end{align}\]
Thus, for the roots to be equal, the difference between the square of the coefficient of x and four times the product of the coefficient of ${{\text{x}}^{2}}$ and constant term must be equal to 0.
Now, the equation given to us is $\left( {{b}^{2}}+{{c}^{2}} \right){{x}^{2}}-2\left( a+b \right)cx+\left( {{c}^{2}}+{{a}^{2}} \right)=0$.
We will apply the same condition for equal roots in our equation.
$\Rightarrow {{\left( -2\left( a+b \right)c \right)}^{2}}=4\left( {{b}^{2}}+{{c}^{2}} \right)\left( {{c}^{2}}+{{a}^{2}} \right)$
Now, we shall simplify the left-hand side.
$\Rightarrow 4{{\left( a+b \right)}^{2}}{{c}^{2}}=4\left( {{b}^{2}}+{{c}^{2}} \right)\left( {{c}^{2}}+{{a}^{2}} \right)$
4 divides on both sides. We will find ${{\left( a+b \right)}^{2}}$
$\Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right){{c}^{2}}=\left( {{b}^{2}}+{{c}^{2}} \right)\left( {{c}^{2}}+{{a}^{2}} \right)$
Now, we shall solve the parenthesis on both sides.
$\Rightarrow {{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+2ab{{c}^{2}}={{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}+{{c}^{4}}+{{c}^{2}}{{a}^{2}}$
Now, we shall cancel the terms common for both sides.
\[\begin{align}
  & \Rightarrow 2ab{{c}^{2}}={{c}^{4}}+{{a}^{2}}{{b}^{2}} \\
 & \Rightarrow {{c}^{4}}+{{a}^{2}}{{b}^{2}}-2ab{{c}^{2}}=0 \\
 & \Rightarrow {{\left( {{c}^{2}}-ab \right)}^{2}}=0 \\
\end{align}\]
Thus, we take square root on both sides.
\[\begin{align}
  & \Rightarrow {{c}^{2}}-ab=0 \\
 & \Rightarrow {{c}^{2}}=ab \\
\end{align}\]
Thus, we can say that a, c, b are in GP.

Note: Similarly, we can also find the conditions for 2 unique solutions and no real solutions. If ${{b}^{2}}-4ac>0$, this means the quadratic equation has 2 unique roots, and if ${{b}^{2}}-4ac<0$, this means the quadratic equation has no real roots.