Question

The value of \[p\] for which the sum of the squares of the roots of \[2{x^2} - 2(p - 2)x - p - 1 = 0\] is least is
(A) \[1\]
(B) \[\dfrac{3}{2}\]
(C) \[2\]
(D) \[ - 1\]

Answer 1

Hint: To solve this problem, we use the concepts and properties of quadratic equations. And one of its properties is, sum of roots property. An equation of kind \[a{x^2} + bx + c = 0\] is a quadratic equation. If its roots are \[\alpha ,\beta \] , then, the sum of these roots is equal to the negative coefficient of \[x\] divided by the coefficient of \[{x^2}\] . That means, \[\alpha + \beta = - \dfrac{b}{a}\] . And similarly, the product of roots is \[\alpha \times \beta = \dfrac{c}{a}\] .
And another property is about the maximum and the minimum values of a quadratic equation.

Complete step by step answer:
The equation is given as \[2{x^2} - 2(p - 2)x - p - 1 = 0\]
Let roots of this equation be \[\alpha \] and \[\beta \] .
So sum of roots is \[\alpha + \beta = - \dfrac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}\]
\[ \Rightarrow \alpha + \beta = - \dfrac{{ - 2(p - 2)}}{2} = p - 2\] -----\[(1)\]
Product of roots is \[\alpha \times \beta = \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}\]
\[ \Rightarrow \alpha \times \beta = \dfrac{{ - p - 1}}{2}\] ------\[(2)\]
Now let us evaluate the value of the sum of squares.
So, the sum of squares is equal to \[{\alpha ^2} + {\beta ^2}\] .
And we know the identity, \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and from this equation we can conclude that \[{a^2} + {b^2} = {(a + b)^2} - 2ab\] .
So similarly, we get \[{\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta \] ------\[(3)\]
And we know the values of \[\alpha + \beta \] and \[\alpha \beta \] from equations \[(1)\] and \[(2)\] .
So, substituting those values in \[(3)\] , we get, \[{\alpha ^2} + {\beta ^2} = {(p - 2)^2} - ( - p - 1)\]
\[ \Rightarrow {\alpha ^2} + {\beta ^2} = {p^2} + 4 - 4p + p + 1\]
\[ \Rightarrow {\alpha ^2} + {\beta ^2} = {p^2} - 3p + 5\]
And it is given that this value is least.
A quadratic equation \[a{x^2} + bx + c = 0\] gains its least value at \[x = - \dfrac{b}{{2a}}\] (if \[a > 0\] ) or gains its maximum value at \[x = - \dfrac{b}{{2a}}\] (if \[a < 0\] ).
So here, we got a quadratic in \[p\] , and the coefficient of \[{p^2}\] is \[1\] which is greater than \[0\] .
So, we get least value at \[p = - \dfrac{{ - 3}}{2}\]
So, the value of \[p\] for which sum of squares of roots is least is \[p = \dfrac{3}{2}\] .

So, the correct answer is “Option B”.

Note:
While taking the coefficient values, the signs should also be included, or else, you will not get the correct values and your solution goes wrong. And also, be perfect with the identities and the properties related to quadratic equations. Also, look whether the coefficient of the quadratic equation that you got is positive or negative, because that can change the whole sum too.