Question

If $a,b,c\in R$ and roots of the equation $a{{x}^{2}}+2bx+c=0$ are real and different, then roots of the equation $\left( {{a}^{2}}+2{{b}^{2}}-ac \right){{x}^{2}}+2b\left( a+c \right)x+\left( 2{{b}^{2}}+{{c}^{2}}-ac \right)=0$ are
(a) real and equal
(b) real and unequal
(c) imaginary
(d) none of these

Answer 1

Hint: In order to find the nature of the roots for any given equation, the best way is to find the discriminant of the given equation. The standard quadratic equation is of the form $a{{x}^{2}}+bx+c=0............(i)$, then the discriminant of the standard quadratic equation is ${{b}^{2}}-4ac$. Now, we can determine the nature of the roots based on the following given conditions:
(i) If ${{b}^{2}}-4ac=0$, then the roots are real, rational and equal.
(ii) If ${{b}^{2}}-4ac>0$ and is also a perfect square, then the roots are real, rational, and distinct.
(iii) If ${{b}^{2}}-4ac>0$ and is also not a perfect square, then the roots are real, irrational, and distinct.
(iv) If ${{b}^{2}}-4ac<0$, then the roots are imaginary.

Complete step-by-step solution:
Here, we are given that $a,b,c\in R$ and roots of the equation $a{{x}^{2}}+2bx+c=0...........(ii)$ are real and different. We have to find the nature of the roots for the equation $\left( {{a}^{2}}+2{{b}^{2}}-ac \right){{x}^{2}}+2b\left( a+c \right)x+\left( 2{{b}^{2}}+{{c}^{2}}-ac \right)=0......................(iii)$.
Now, on comparing the given equation (i) with the standard quadratic equation (iii), we get,
$a=\left( {{a}^{2}}+2{{b}^{2}}-ac \right)$
$b=2b\left( a+c \right)$
$c=\left( 2{{b}^{2}}+{{c}^{2}}-ac \right)$
Similarly, on comparing the given equation (i) with the standard quadratic equation (ii), we get,
$a=a$
$b=2b$
$c=c$
The discriminant of the standard quadratic equation is given by ${{b}^{2}}-4ac$. Based on the value of discriminant, we can easily determine the nature of the roots of the equation.
Let ${{\text{D}}_{\text{1}}}$ and ${{\text{D}}_{\text{2}}}$ denote the discriminant of the equation (ii) and (iii). Now, the discriminant of equation (i) is given by –
$\Rightarrow {{\text{D}}_{\text{1}}}={{b}^{2}}-4ac$
$\Rightarrow {{\text{D}}_{\text{1}}}={{b}^{2}}-4ac............(iv)$
Now, replacing the value of a, b, c in equation (iv), we get,
$\Rightarrow {{\text{D}}_{\text{1}}}={{\left( 2b \right)}^{2}}-4ac$
$\Rightarrow {{\text{D}}_{\text{1}}}=4{{b}^{2}}-4ac$
$\Rightarrow {{\text{D}}_{\text{1}}}=4\left( {{b}^{2}}-ac \right)..............(v)$
Since, the roots of the equation (i) are real and distinct, so, the condition of discriminant will be as follows:
$\Rightarrow {{\text{D}}_{\text{1}}}>0.............(vi)$
Hence, from equation (v) and (vi), we can write,
$\Rightarrow 4\left( {{b}^{2}}-ac \right)>0$
$\Rightarrow {{b}^{2}}-ac>0$
Now, let $k={{b}^{2}}-ac>0............(vi)$
Then, Finding the discriminant of equation (iii) using the values of a, b, c obtained above after comparing equation (iii) with standard quadratic equation, we get,
$\Rightarrow {{\text{D}}_{\text{2}}}={{\left( 2b\left( a+c \right) \right)}^{2}}-4\left( {{a}^{2}}+2{{b}^{2}}-ac \right)\left( 2{{b}^{2}}+{{c}^{2}}-ac \right)$
Here in the above term to separate 4 from the both the equations we have to square the
${{\left( 2b\left( a+c \right) \right)}^{2}}$ then we will get, $4[{{b}^{2}}{{\left( a+c \right)}^{2}}]$ this way we will solve the next step
$\begin{align}
  & \Rightarrow {{\text{D}}_{\text{2}}}=\left( {{(2)}^{2}}{{b}^{2}}{{\left( a+c \right)}^{2}} \right)-4\left( {{a}^{2}}+2{{b}^{2}}-ac \right)\left( 2{{b}^{2}}+{{c}^{2}}-ac \right) \\
 & \Rightarrow {{\text{D}}_{\text{2}}}=\left( (4){{b}^{2}}{{\left( a+c \right)}^{2}} \right)-(4)\left( {{a}^{2}}+2{{b}^{2}}-ac \right)\left( 2{{b}^{2}}+{{c}^{2}}-ac \right) \\
 & \\
\end{align}$
$\Rightarrow {{\text{D}}_{\text{2}}}=4\left( {{b}^{2}}{{\left( a+c \right)}^{2}}-\left( {{a}^{2}}+2{{b}^{2}}-ac \right)\left( 2{{b}^{2}}+{{c}^{2}}-ac \right) \right)$
In the above equation we will substitute $\Rightarrow {{b}^{2}}=ac+k$ from equation number (vii) then we will get the term as,
$\begin{align}
  & \Rightarrow {{\text{D}}_{\text{2}}}=4\left( \left( ac+k \right){{\left( a+c \right)}^{2}}-\left( {{a}^{2}}+2\left( ac+k \right)-ac \right)\left( 2\left( ac+k \right)+{{c}^{2}}-ac \right) \right) \\
 & \Rightarrow {{\text{D}}_{\text{2}}}=4\left( \left( ac+k \right){{\left( a+c \right)}^{2}}-\left( {{a}^{2}}+2ac+2k-ac \right)\left( 2ac+2k+{{c}^{2}}-ac \right) \right) \\
\end{align}$
so here $2ac-ac=ac$ we will get,
$\Rightarrow {{\text{D}}_{\text{2}}}=4\left( \left( ac+k \right){{\left( a+c \right)}^{2}}-\left( {{a}^{2}}+2k+ac \right)\left( 2k+{{c}^{2}}+ac \right) \right)$
$\Rightarrow {{\text{D}}_{\text{2}}}=4\left( ac{{\left( a+c \right)}^{2}}+k{{\left( a+c \right)}^{2}}-\left( {{a}^{2}}\left( 2k+{{c}^{2}}+ac \right)+2k\left( 2k+{{c}^{2}}+ac \right)+ac\left( 2k+{{c}^{2}}+ac \right) \right) \right)$
$\Rightarrow {{\text{D}}_{\text{2}}}=4\left( ac{{\left( a+c \right)}^{2}}+k{{\left( a+c \right)}^{2}}-\left( {{a}^{2}}.ac+ac.{{c}^{2}}+{{a}^{2}}.{{c}^{2}}+{{a}^{2}}.{{c}^{2}}+{{a}^{2}}.2k+2k.ac+2k.ac+2k.{{c}^{2}}+2k.2k \right) \right)$
so, in the above equation we will be applying the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ then we will get,
$\Rightarrow {{\text{D}}_{\text{2}}}=4\left( ac{{\left( a+c \right)}^{2}}+k{{\left( a+c \right)}^{2}}-ac{{\left( a+c \right)}^{2}}-2k{{\left( a+c \right)}^{2}}-4{{k}^{2}} \right)$
After subtracting the terms, we will get are,
$\therefore {{\text{D}}_{\text{2}}}=4\left( -k{{\left( a+c \right)}^{2}}-4{{k}^{2}} \right)< 0$
Thus, since the discriminant is a negative value, the nature of the roots is imaginary.
Hence, the correct option is an option (c).

Note: Normally, students make mistakes while finding the discriminant of the quadratic equation. Besides, the student gets confused about the condition that leads to the particular nature of roots. Students should always remember the condition of the discriminant that determines the nature of the roots. Moreover, students should be careful while replacing the value of ${{b}^{2}}-ac$ in the discriminant of the equation whose roots are given.