Question

If m is chosen in the quadratic equation $\left( {{m}^{2}}+1 \right){{x}^{2}}-3x+{{\left( {{m}^{2}}+1 \right)}^{2}}=0$ such that the sum of its roots is greatest, then find the absolute difference of the cubes of its roots?
(a) $8\sqrt{3}$,
(b) $4\sqrt{3}$,
(c) $10\sqrt{5}$,
(d) $8\sqrt{5}$.

Answer 1

We start solving the problem by assigning variables for the sum and product of the roots of the given quadratic equation. We find the sum of the roots and check the value of m at which the sum becomes maximum. After finding the value of m we calculate the sum and product of the roots. We then make necessary calculations to get the required result.

Complete step by step answer:
According to the problem, we have a quadratic equation $\left( {{m}^{2}}+1 \right){{x}^{2}}-3x+{{\left( {{m}^{2}}+1 \right)}^{2}}=0$ and value m is to be chosen such that the sum of its roots is greatest. We need to find the absolute difference of the cubes of its roots. Let us assume the sum of its roots be s and the product of its roots be p.
We know that the sum of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$. We use this result to find the sum of roots of the quadratic equation $\left( {{m}^{2}}+1 \right){{x}^{2}}-3x+{{\left( {{m}^{2}}+1 \right)}^{2}}=0$.
We get the sum of the roots $s=\dfrac{-\left( -3 \right)}{{{m}^{2}}+1}$.
$\Rightarrow s=\dfrac{3}{{{m}^{2}}+1}$ ---(1).
We need to find the value of m such that the value of s is maximum. From equation (1), we can see that the numerator of s is positive. In order to get the maximum value of s, we need the denominator to be minimum.
We have a denominator as ${{m}^{2}}+1$ and we know that the square of a positive or negative real number is greater than 0. So, we get the value of ${{m}^{2}}$ as 0 if the value of m is 0 which will be the minimum value for ${{m}^{2}}$. We get the minimum value of m as 0.
Let us substitute the $m=0$ in the given quadratic equation $\left( {{m}^{2}}+1 \right){{x}^{2}}-3x+{{\left( {{m}^{2}}+1 \right)}^{2}}=0$.
$\Rightarrow \left( {{0}^{2}}+1 \right){{x}^{2}}-3x+{{\left( {{0}^{2}}+1 \right)}^{2}}=0$.
$\Rightarrow \left( 0+1 \right){{x}^{2}}-3x+{{\left( 0+1 \right)}^{2}}=0$.
$\Rightarrow \left( 1 \right){{x}^{2}}-3x+{{\left( 1 \right)}^{2}}=0$.
$\Rightarrow {{x}^{2}}-3x+1=0$ ---(2).
Let us assume p and q be the roots of equation (2).
We know that the sum and product of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$. We apply this result in equation (2).
So, we get $p+q=\dfrac{-\left( -3 \right)}{1}$ and $pq=\dfrac{1}{1}$.
$\Rightarrow p+q=3$ and $pq=1$.
Now we need to find the value of ${{p}^{3}}-{{q}^{3}}$.
We know that ${{p}^{3}}-{{q}^{3}}=\left( p-q \right)\times \left( {{p}^{2}}+{{q}^{2}}+pq \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{{{\left( p-q \right)}^{2}}} \right)\times \left( {{p}^{2}}+{{q}^{2}}+2pq-pq \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{{{p}^{2}}+{{q}^{2}}-2pq} \right)\times \left( {{\left( p+q \right)}^{2}}-pq \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{{{p}^{2}}+{{q}^{2}}+2pq-4pq} \right)\times \left( {{\left( p+q \right)}^{2}}-pq \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{{{\left( p+q \right)}^{2}}-4pq} \right)\times \left( {{\left( p+q \right)}^{2}}-pq \right)$.
Let us substitute $p+q=3$ and $pq=1$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)} \right)\times \left( {{\left( 3 \right)}^{2}}-1 \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{9-4} \right)\times \left( 9-1 \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=\left( \sqrt{5} \right)\times \left( 8 \right)$.
$\Rightarrow {{p}^{3}}-{{q}^{3}}=8\sqrt{5}$.
We have found the absolute value of the difference of cubes of the roots as $8\sqrt{5}$.
∴ The absolute value of the difference of cubes of the roots is $8\sqrt{5}$.
The correct option for the given problem is (d).

Note:
We can also find the value of ${{p}^{3}}-{{q}^{3}}$ by finding the value of $p-q$ first and later use it. We should know that any square of a real number except 0 is greater than 0. We should not make mistakes while performing addition and subtraction operations. Similarly, we can expect problems to find the roots of the given quadratic equation after finding the value of m.