Question

Find the value of 'a' such that the sum of the squares of the roots of the equation x2 - (a-2)x - (a+1) = 0 is least.​

Answer 1

sum of roots of quadratic equation

= -(co-efficient of x)/(co-efficient of x²)

= a-2/1

= a-2

product of roots= -(a+1)

let's assume the two roots of the equation to be 'm' and 'n'

then

(m+n)²=m²+n²+2mn

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

$\sf{Let \: \alpha \: and \: \beta \: be \: the \: roots \: of \: the \: equation}$

$\sf{\therefore \: \alpha + \beta = \frac{ - b}{a}}$

$\sf { \implies \: \alpha + \beta = a - 2}$

and

$\sf {\implies \: \alpha \beta = \frac{c}{a} }$

$\sf{ \implies \: \alpha \beta = - (a + 1)}$

Now,

$\implies \: \sf{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }$

Putting the values we get,

$= \sf{\: {(a - 2)}^{2} + 2( a + 1)}$

$= \: \sf{({a}^{2} + 4 - 4a) + 2a + 2}$

$= \sf{{a}^{2} + 1 + 3 - 2a + 2}$

$= \: \sf{ ({a}^{2} + 1 - 2a )+ 3 + 2 }$

$= \sf {{(a - 1)}^{2} + 5}$

Now,

${ \alpha }^{2} + { \beta }^{2}$ will be minimum only when

$\implies \: \sf{ {(a - 1)}^{2} = 0}$

$\sf{ \implies \: (a - 1) = 0}$

$\sf{\implies \: a = 1}$

$\boxed{ \sf{Hence, \: a = 1}}$