Question

For what value of $\lambda $ the sum of the squares of the root of ${{x}^{2}}+\left( 2+\lambda \right)x-\dfrac{1}{2}\left( 1+\lambda \right)=0$ is minimum.

Answer 1

Hint:This question involves a simple concept of quadratic equation and its roots. Here in given equation we have to get sum of roots and products of roots and then by using the formula –
${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $
We can get ${{\left( \alpha +\beta \right)}^{2}}$in terms of$\lambda $, then we get quadratic equation in terms of$\lambda $. We can then calculate $\lambda $for minimum value of the quadratic by making perfect square.

Complete step by step answer:
(i) Let a quadratic equation $a{{x}^{2}}+bx+c=0$ has roots $\alpha $ and$\beta $.
Then,
Sum of roots $=-\dfrac{b}{a}$
Product of roots $=\dfrac{c}{a}$

(ii) For making sum of the squares of roots minimum
$a{{x}^{2}}+bx+c=0$
By taking $'a'$ common from the equation,
$a\left( {{x}^{2}}+\dfrac{bx}{a}+\dfrac{c}{a} \right)=0$
$\because a\ne 0$
$\Rightarrow {{x}^{2}}+\dfrac{bx}{a}+\dfrac{c}{a}=0$
$\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}-\dfrac{{{b}^{2}}}{4{{a}^{2}}}=0$

For making sum of the squares of roots minimum, perfect square should be zero.
So, $x+\dfrac{b}{2a}=0$
$\Rightarrow x=-\dfrac{b}{2a}$
For$x=-\dfrac{b}{2a}$, equation $a{{x}^{2}}+bx+c=0$will be minimum.
Let minimum value be$y$.
So,
$y=\left( \dfrac{c}{a}-\dfrac{{{b}^{2}}}{4{{a}^{2}}} \right)a$
$\Rightarrow y=\left( \dfrac{4ac-{{b}^{2}}}{4{{a}^{2}}} \right)a$
$\Rightarrow y=\left( \dfrac{-D}{4a} \right)$

Now, given equation is –
${{x}^{2}}+\left( 2+\lambda \right)x-\dfrac{1}{2}\left( 1+\lambda \right)=0$
Let this equation has roots $\alpha $ and$\beta $. Now by comparing this equation with$a{{x}^{2}}+bx+c=0$, we have
$a=1$ , $\Rightarrow b=\left( 2+\lambda \right)$ , $\Rightarrow c=-\dfrac{1}{2}\left( 1+\lambda \right)$

Sum of roots $=-\dfrac{b}{a}$
$\Rightarrow \alpha +\beta =-\left( \lambda +2 \right)$
Product of roots $=\dfrac{c}{a}$
\[\Rightarrow \alpha \beta =-\dfrac{1}{2}\left( \lambda +1 \right)\]

Now according to the question, we have to get the value of${{\alpha }^{2}}+{{\beta }^{2}}$.
\[{{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \]
Let us put the values in this equation.
\[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left\{ -\left( \lambda +2 \right) \right\}}^{2}}-2\left\{ -\dfrac{1}{2}\left( \lambda +1 \right) \right\}\]
\[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+4\lambda +4+\left( \lambda +1 \right)\]
\[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+5\lambda +5\]

Now we have to calculate $\lambda $ for minimum\[{{\alpha }^{2}}+{{\beta }^{2}}\], so we will make perfect square.
\[{{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+5\lambda +5\]
\[={{\left( \lambda +\dfrac{5}{2} \right)}^{2}}+5-\dfrac{25}{4}\]
\[={{\left( \lambda +\dfrac{5}{2} \right)}^{2}}-\dfrac{5}{4}\]
For minimum \[{{\alpha }^{2}}+{{\beta }^{2}}\], square should be zero.
\[\Rightarrow {{\left( \lambda +\dfrac{5}{2} \right)}^{2}}=0\]
\[\Rightarrow \lambda +\dfrac{5}{2}=0\]
\[\Rightarrow \lambda =-\dfrac{5}{2}\]

Hence, for\[\lambda =-\dfrac{5}{2}\], \[{{\alpha }^{2}}+{{\beta }^{2}}\]will be minimum.

Note:
In this question, while making square of the equation, we have to take care that the term of $x$ will be compared with the term $2ab$ of square of${{\left( a+b \right)}^{2}}=\left( {{a}^{2}}+2ab+{{b}^{2}} \right)$.
Let the equation is${{x}^{2}}+px+q$, and we compare this equation with ${{\left( a+b \right)}^{2}}=\left( {{a}^{2}}+2ab+{{b}^{2}} \right)$
$\Rightarrow a=x$ , $\Rightarrow 2b=p$ $\Rightarrow b=\dfrac{p}{2}$
So,
${{x}^{2}}+px+q={{\left( x+\dfrac{p}{2} \right)}^{2}}+q-\dfrac{{{p}^{2}}}{4}$

(ii) In the formula${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $, sign negative is important. Students make mistake that they take positive instead of negative.