Question

If the sum of the squares of the roots of the equation \[{x^2} - (a - 2)x - (a + 1) = 0\] is least, then the value of a is
A) -1
B) 1
C) 2
D) -2

Answer 1

Hint:
At first, we will find the formula of the sum of the squares of the roots for the general quadratic equation. Then, we will use the formula and determine the sum of the squares of the roots of the given equation. For finding the least value of the sum of the squares of the roots of the equation, we will take the derivative of the equation obtained. On simplification, we will get the value of \[a\].

Complete step by step solution:
The given equation is \[{x^2} - (a - 2)x - (a + 1) = 0\] …(1)
The general equation for quadratic equation is
 \[ \Rightarrow a{x^2} + bx + c = 0\]
Let \[\alpha \] and \[\beta \] are its roots.
Then, the sum of roots is
 \[ \Rightarrow \alpha + \beta = - \dfrac{b}{a}\]
And the product of roots
 \[ \Rightarrow \alpha \beta = \dfrac{c}{a}\]
Square of \[\alpha + \beta \] gives
 \[ \Rightarrow {(\alpha + \beta )^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta \]
Taking \[{\alpha ^2} + {\beta ^2}\] on left hand side and the remaining terms on the right-hand side.
 \[ \Rightarrow {\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta \] ….(2)
Given, \[{x^2} - (a - 2)x - (a + 1) = 0\] ….(3)
If the roots of this equation are \[\gamma \] and \[\delta \]
Then the sum of roots is
 \[ \Rightarrow \gamma + \delta = a - 2\] ….(4)
And the product of roots is
 \[ \Rightarrow \gamma \delta = - (a + 1)\] ….(5)
The sum of square of roots can be written as shown in equation (2)
 \[ \Rightarrow ({\gamma ^2} + {\delta ^2}) = {(\gamma + \delta )^2} - 2\gamma \delta \] ….(6)
Substituting the value of equation (4) and (5) in equation (6)
 \[ \Rightarrow ({\gamma ^2} + {\delta ^2}) = {(a - 2)^2} + 2(a + 1)\]
On simplification we get
 \[ = {a^2} + 4 - 4a + 2a + 2\]
On adding like terms, we get,
 \[ = {a^2} - 2a + 6\]
 \[ \Rightarrow ({\gamma ^2} + {\delta ^2}) = {a^2} - 2a + 6\] ….(7)
Since the value of the sum of squares of the roots of the equation is least.
Therefore, to find the least value, we will take the derivative of it.
 \[ \Rightarrow \dfrac{{d({a^2} - 2a + 6)}}{{da}} = 0\]
The derivative of \[{a^2} - 2a + 6\] is
 \[ \Rightarrow 2a - 2 = 0\]
On dividing by 2 we get,
 \[ \Rightarrow a - 1 = 0\]
On adding 1 to both the sides we get,
 \[ \Rightarrow a = 1\]
Therefore, if the sum of squares of the roots of the equation \[{x^2} - (a - 2)x - (a + 1) = 0\] is least, then the value of a is 1.

Hence option B is correct.

Note:
Consider a quadratic equation \[a{x^2} + bx + c = 0\] , and let \[\alpha \] and \[\beta \] are its roots.
Then, the sum of the roots is given by
 \[ \Rightarrow \alpha + \beta = - \dfrac{b}{a}\]
And the product of the roots is given by
 \[ \Rightarrow \alpha \beta = \dfrac{c}{a}\]
And the roots of the quadratic equation are given by
 \[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]