Question

For what value of $\lambda$ the sum of the squares of the root of $x^2+(2+\lambda )x-{\displaystyle \frac{1}{2}}(1+\lambda )=0$ 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 –
${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$
We can get ${(\alpha +\beta )}^2$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 $=-{\displaystyle \frac{b}{a}}$
Product of roots $={\displaystyle \frac{c}{a}}$

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

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

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

Sum of roots $=-{\displaystyle \frac{b}{a}}$
$\Rightarrow \alpha +\beta =-(\lambda +2)$
Product of roots $={\displaystyle \frac{c}{a}}$
$$\Rightarrow \alpha \beta =-{\displaystyle \frac{1}{2}}(\lambda +1)$$

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

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

Hence, for$$\lambda =-{\displaystyle \frac{5}{2}}$$, $${\alpha }^2+{\beta }^2$$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${(a+b)}^2=(a^2+2ab+b^2)$.
Let the equation is$x^2+px+q$, and we compare this equation with ${(a+b)}^2=(a^2+2ab+b^2)$
$\Rightarrow a=x$ , $\Rightarrow 2b=p$ $\Rightarrow b={\displaystyle \frac{p}{2}}$
So,
$x^2+px+q={(x+{\displaystyle \frac{p}{2}})}^2+q-{\displaystyle \frac{p^2}{4}}$

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