For what value of a the sum of squares of the roots of the equation x2-(a-2) x-(a+1)=0 is least?

Question

For what value of a the sum of squares of the roots of the equation x2-(a-2) x-(a+1)=0 is least? (A) 1 (B) 2 (C) -1 (D) -2

Tardigrade Admin · Accepted Answer

Let α, β be the roots of the equation. Therefore, α+β=a-2 and α β=-(a+1) Now,α2+β2 =(α+β)2-2 α β =(a-2)2+2(a+1) =(a-1)2+5 Therefore, α2+β2 will be minimum, if (a-1)2=0, i.e., a=1.

Q. For what value of $a$ the sum of squares of the roots of the equation $x^{2} - (a - 2) x - (a + 1) = 0$ is least?

1

2

$- 1$

$- 2$

Let $α, β$ be the roots of the equation.
Therefore, $α + β = a - 2$ and $α β = - (a + 1)$
Now, $α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= (a - 2)^{2} + 2 (a + 1)$
$= (a - 1)^{2} + 5$
Therefore, $α^{2} + β^{2}$ will be minimum, if $(a - 1)^{2} = 0$ , i.e., $a = 1$ .

Q. For what value of a the sum of squares of the roots of the equation x2−(a−2)x−(a+1)=0 is least?

1

2

−1

−2

Let α,β be the roots of the equation. Therefore, α+β=a−2 and αβ=−(a+1) Now,α2+β2=(α+β)2−2αβ =(a−2)2+2(a+1) =(a−1)2+5 Therefore, α2+β2 will be minimum, if (a−1)2=0, i.e., a=1.

Q. For what value of $a$ the sum of squares of the roots of the equation $x^{2} - (a - 2) x - (a + 1) = 0$ is least?

$- 1$

$- 2$

Let $α, β$ be the roots of the equation.
Therefore, $α + β = a - 2$ and $α β = - (a + 1)$
Now, $α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= (a - 2)^{2} + 2 (a + 1)$
$= (a - 1)^{2} + 5$
Therefore, $α^{2} + β^{2}$ will be minimum, if $(a - 1)^{2} = 0$ , i.e., $a = 1$ .