Question

The equation \[{{x}^{2}}+bx+c=0\] has distinct roots. If 2 is subtracted from each root then the resultant roots are reciprocal from original roots. Find the value of ${{b}^{2}}+{{c}^{2}}$ .

Answer 1

Hint: We are given with the condition that if 2 is subtracted from root the resultant is reciprocal of original roots. Hence from this condition we can find two quadratic equations. Hence we will find the values of two roots. Now we know that for a quadratic equation $a{{x}^{2}}+bx+c=0$ . We have sum of roots = $-\dfrac{b}{a}$ and product of roots = $\dfrac{c}{a}$. Hence we have now relation between roots and b, c. From this we can easily find the value of ${{b}^{2}}+{{c}^{2}}$ .

Complete step by step answer:
Now the given quadratic equation is \[{{x}^{2}}+bx+c=0\].
Let α and β be the roots of the equation.
Then from the given condition we know that
$\alpha -2=\dfrac{1}{\alpha }$ and $\beta -2=\dfrac{1}{\beta }$
Now let us first consider the equation $\alpha -2=\dfrac{1}{\alpha }$
$\begin{align}
  & \Rightarrow {{\alpha }^{2}}-2\alpha =1 \\
 & \therefore {{\alpha }^{2}}-2\alpha -1=0 \\
\end{align}$
Now this is again a quadratic in α
We know that the roots of equation $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence for equation ${{\alpha }^{2}}-2\alpha -1=0$ we have
$\begin{align}
  & \alpha =\dfrac{-(-2)\pm \sqrt{{{(-2)}^{2}}-4(1)(-1)}}{2(1)} \\
 & \Rightarrow \alpha =\dfrac{2\pm \sqrt{4+4}}{2}=\dfrac{2\pm \sqrt{4\times 2}}{2} \\
 & \therefore \alpha =\dfrac{2\pm 2\sqrt{2}}{2} \\
\end{align}$
Hence we get $\alpha =1\pm \sqrt{2}.............(1)$
Now consider $\beta -2=\dfrac{1}{\beta }$
$\begin{align}
  & \Rightarrow {{\beta }^{2}}-2=1 \\
 & \therefore {{\beta }^{2}}-2\beta -1=0 \\
\end{align}$
Now this is again a quadratic in $\beta $
We know that the roots of equation $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence for equation ${{\beta }^{2}}-2\beta -1=0$ we have
\[\begin{align}
  & \beta =\dfrac{-(-2)\pm \sqrt{{{(-2)}^{2}}-4(1)(-1)}}{2(1)} \\
 & \Rightarrow \beta =\dfrac{2\pm \sqrt{4+4}}{2}=\dfrac{2\pm \sqrt{4\times 2}}{2} \\
 & \therefore \beta =\dfrac{2\pm 2\sqrt{2}}{2} \\
\end{align}\]
Hence we get $\beta =1\pm \sqrt{2}..................(2)$
Now we are given that the roots are different using this in equation (1) and (2) we get
The two roots must be $1+\sqrt{2}$ and $1-\sqrt{2}$
Hence let us take $\alpha =1+\sqrt{2}$ and $\beta =1-\sqrt{2}$……………………… (3)
.Now we know that for a quadratic equation $a{{x}^{2}}+bx+c=0$ . We have sum of roots = $-\dfrac{b}{a}$ and product of roots = $\dfrac{c}{a}$.
Now our quadratic equation is \[{{x}^{2}}+bx+c=0\]
Hence from (3) we get
$\begin{align}
  & 1+\sqrt{2}+1-\sqrt{2}=b \\
 & \therefore b=2.......................(4) \\
\end{align}$
And
$\begin{align}
  & \left( 1+\sqrt{2} \right)\left( 1-\sqrt{2} \right)=c \\
 & \Rightarrow c={{1}^{2}}-{{\left( \sqrt{2} \right)}^{2}} \\
 & \Rightarrow c=1-2 \\
 & \therefore c=-1.....................(5) \\
\end{align}$
Now from equation (4) and equation (5) we get
${{b}^{2}}+{{c}^{2}}={{2}^{2}}+{{\left( -1 \right)}^{2}}=2+1=3$
Hence ${{b}^{2}}+{{c}^{2}}=3$.

Note:
Now while taking roots note that the roots are different and hence we have not taken conditions such as $\alpha =1+\sqrt{2},\beta =1+\sqrt{2}$ and $\alpha =1-\sqrt{2},\beta =1-\sqrt{2}$
Also here we have considered $\alpha =1+\sqrt{2}$ and $\beta =1-\sqrt{2}$ it doesn’t matter if we take $\beta =1+\sqrt{2}$ and $\alpha =1-\sqrt{2}$ . as in the end the sum of roots and product of roots will still come same.