Question

The value of λ such that sum of the squares of the roots of the quadratic equation, ${{x}^{2}}+(3-\lambda )x+2=\lambda $ has the least value is:
(a) 2
(b) $\dfrac{4}{9}$
(c) $\dfrac{15}{8}$
(d) 1

Answer 1

Hint: To solve the above question, we are using two equations; the sum of roots and product of roots of a general quadratic equation. By further solving equations we get the value of λ which is required. The general quadratic equation is $a{{x}^{2}}+b x + c=0$ , so the sum of roots will be $\dfrac{-b}{a}$ and the product of roots will be $\dfrac{c}{a}$.

Complete step by step answer:
The quadratic equation is ${{x}^{2}}+\left( 3-\lambda \right)x+2=\lambda $. We need to convert this to the format of general quadratic equation i.e. $a{{x}^{2}}+bx+c=0$. Therefore, the given equation becomes as ${{x}^{2}}-\left( 3-\lambda \right)x+2-\lambda =0$
Now, we know that, sum of roots of a general quadratic equation is $\dfrac{-b}{a}$ and the product of roots of quadratic equation$a{{x}^{2}}+bx+c=0$ is $\dfrac{c}{a}$.
Consider the roots of the quadratic equation as $\alpha $, $\beta $.
 Hence, we get two equations as
$\alpha +\beta =\dfrac{-b}{a}=\dfrac{-\left( 3-\lambda \right)}{1}=\lambda -3......\left( 1 \right)$
$\Rightarrow \alpha \beta =\dfrac{c}{a}=\dfrac{2-\lambda }{1}=2-\lambda ......\left( 2 \right)$
According, to the given condition we need to find the minimum value of ${{(\alpha )}^{2}}+{{(\beta )}^{2}}$.
Now, to form the required expression, form an equation as,
$\Rightarrow {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}$
After further solving we get,
$\begin{align}
  & {{\left( 3-\lambda \right)}^{2}}-2\left( 2-\lambda \right)={{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}} \\
 & \Rightarrow {{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}=\left( {{\lambda }^{2}}-6\lambda +9 \right)-2\left( 2-\lambda \right) \\
 & \Rightarrow {{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}=\left( {{\lambda }^{2}}-6\lambda +9+2\lambda -4 \right) \\
 &\Rightarrow {{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}=\left( {{\lambda }^{2}}-4\lambda -5 \right) \\
\end{align}$
For this expression ${{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}$to be least, we need to minimise $\left( {{\lambda }^{2}}-4\lambda -5 \right)$.
Now, we know that if we have a function f(x), then if $\dfrac{d}{d\lambda }f(x)$ exists and $\dfrac{{{d}^{2}}}{d{{\lambda }^{2}}}f(x) > 0$, then $\dfrac{d}{d\lambda }f(x)=0$, gives the values of x, where f(x) is minimum.
So, now minimise $\left( {{\lambda }^{2}}-4\lambda -5 \right)$ i.e. differentiate this equation with respect to λ and then minimise the equation that we get.
$\Rightarrow \dfrac{d}{d\lambda }\left( {{\lambda }^{2}}-4\lambda -5 \right)=2\lambda -4$
Again differentiating, we get
$\Rightarrow \dfrac{d}{d\lambda }\left( 2\lambda -4 \right)=2$
So, as $\dfrac{{{d}^{2}}}{d{{\lambda }^{2}}}({{\lambda }^{2}}-4\lambda -5)=2>0$
So, to find minimum value of $\left( {{\lambda }^{2}}-4\lambda -5 \right)$
$\Rightarrow \dfrac{d}{d\lambda }\left( {{\lambda }^{2}}-4\lambda -5 \right)=0$
$\Rightarrow 2\lambda -4=0$
$\Rightarrow 2\lambda =4$
$\Rightarrow \lambda =2$
Hence, at λ = 2 this expression has minimum value.

So, the correct answer is “Option A”.

Note: You can most probably make mistakes while forming the equations, i.e. the equations of sum of roots of a quadratic equation and product of roots of equation. While forming ${{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\left( \alpha \right)}^{2}}+{{\left( \beta \right)}^{2}}$ this equation be cautious to avoid any mistakes. You can also get mistaken while forming the minimising expression.