Aspire Faculty ID #18147 · Topic: NIMCET 2016 · Just now
NIMCET 2016

The value of a, for which the sum of the square of the roots of the equation $x^2 - (a - 2)x - (a +1) = 0$, assumes the least value is

Solution

The given equation is: 
$x^2 - (a - 2)x - (a + 1) = 0$ 

Let the roots be $\alpha$ and $\beta$. 
$\alpha + \beta = a - 2$ 
$\alpha\beta = -(a + 1)$ 
 Let $f(a)$ represent the sum of the squares of the roots: 
$f(a) = \alpha^2 + \beta^2$ 
$f(a) = (\alpha + \beta)^2 - 2\alpha\beta$ 
 Substitute the expressions: $f(a) = (a - 2)^2 - 2\big(-(a + 1)\big)$ 
$f(a) = (a^2 - 4a + 4) + (2a + 2)$ 
$f(a) = a^2 - 2a + 6$
for mininum 
$f'(a)=0$
$2a-2=0$
$a=1$
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