Question

If p and q are the roots of the quadratic equation \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha =1\], then the minimum value of $\left( {{p}^{2}}+{{q}^{2}} \right)$ is equal to:
(a) 2
(b) 3
(c) 5
(d) 6

Answer 1

Hint: We start solving the problem by writing the given quadratic equation into the standard form $a{{x}^{2}}+bx+c=0$. We use the fact that sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ to find the sum and product of roots of \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha =1\]. We then square the obtained sum of the roots and substitute the value of the product of roots obtained in it. We then use the fact that the minimum value of any square is 0 and make subsequent calculations to find the required value.

Complete step-by-step answer:
According to the problem, we have a quadratic equation \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha =1\] which has roots p and q. We need to find the minimum value of $\left( {{p}^{2}}+{{q}^{2}} \right)$.
Let us convert the given quadratic equation \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha =1\] into the standard form $a{{x}^{2}}+bx+c=0$.
So, we have the quadratic equation \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha -1=0\] ---(1).
We know that sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$. We use this results in equation (1).
So, the sum of roots of the quadratic equation \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha -1=0\] is $p+q=\dfrac{-\left( -\left( \alpha -2 \right) \right)}{1}$.
$\Rightarrow p+q=\alpha -2$ ---(2).
So, the product of roots of the quadratic equation \[{{x}^{2}}-\left( \alpha -2 \right)x-\alpha -1=0\] is $pq=\dfrac{-\alpha -1}{1}$.
$\Rightarrow pq=-\alpha -1$ ---(3).
Let us square on both sides in equation (2).
\[\Rightarrow {{\left( p+q \right)}^{2}}={{\left( \alpha -2 \right)}^{2}}\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}+2pq={{\alpha }^{2}}-4\alpha +4\].
From equation (3) we get,
\[\Rightarrow {{p}^{2}}+{{q}^{2}}+2\left( -\alpha -1 \right)={{\alpha }^{2}}-4\alpha +4\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}-2\alpha -2={{\alpha }^{2}}-4\alpha +4\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}={{\alpha }^{2}}-4\alpha +4+2\alpha +2\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}={{\alpha }^{2}}-2\alpha +6\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}={{\alpha }^{2}}-2\alpha +1+5\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}={{\left( \alpha -1 \right)}^{2}}+5\] ---(4).
We know that the value of the square of any real number is greater than or equal to zero $\left( \ge 0 \right)$. This gives us the minimum value of any square as 0.
So, the minimum value of \[{{\left( \alpha -1 \right)}^{2}}\] is 0. So, let us substitute this value in equation (4) to get the minimum value of \[{{p}^{2}}+{{q}^{2}}\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}=0+5\].
\[\Rightarrow {{p}^{2}}+{{q}^{2}}=5\].
So, we have found the minimum value of \[{{p}^{2}}+{{q}^{2}}\] as 5.
∴ The minimum value of \[{{p}^{2}}+{{q}^{2}}\] as 5.

So, the correct answer is “Option c”.

Note: We can alternatively solve for the minimum value of \[{{p}^{2}}+{{q}^{2}}\] as follows:
Let us assume \[{{p}^{2}}+{{q}^{2}}\] as a function of $\alpha $.
So, we have $f\left( \alpha \right)={{\alpha }^{2}}-2\alpha +6$.
Let's find the roots for ${{f}^{'}}\left( \alpha \right)=0$.
$\Rightarrow \dfrac{d}{d\alpha }\left( {{\alpha }^{2}}-2\alpha +6 \right)=0$.
$\Rightarrow 2\alpha -2=0$.
$\Rightarrow 2\alpha =2$.
$\Rightarrow \alpha =1$.
We know that if the value of the function ${{f}^{''}}\left( \alpha \right)$ is greater than zero at $\alpha =1$, then the function $f\left( \alpha \right)$ has minimum at $\alpha =1$.
$\Rightarrow {{f}^{''}}\left( \alpha \right)=\dfrac{d}{d\alpha }\left( 2\alpha -4 \right)$.
$\Rightarrow {{f}^{''}}\left( \alpha \right)=2$.
$\Rightarrow {{f}^{''}}\left( 1 \right)=2>0$.
So, we have the function minimum value at $\alpha =1$ for the function $f\left( \alpha \right)$.
So, the minimum value of $f\left( \alpha \right)$ is $f\left( 1 \right)$.
$\Rightarrow f\left( 1 \right)={{1}^{2}}-2\left( 1 \right)+6$.
$\Rightarrow f\left( 1 \right)=1-2+6$.
$\Rightarrow f\left( 1 \right)=5$.
The minimum value of \[{{p}^{2}}+{{q}^{2}}\] is 5.