If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13, then find that equation

Question

If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13, then find that equation (A) x2+x-6=0 (B) x2-x+6=0 (C) x2-x-6=0 (D) x2+x+6=0

Tardigrade Admin · Accepted Answer

Let the roots of the quadratic equation are

α

and

β

then it is given that

α + β = 1

and

α^{2} + β^{2} = 13

∴ α β = \frac{1}{2} [(α + β)^{2} - (α^{2} + β^{2})]

= \frac{1}{2} [1 - 13] = - 6

So, equation of required quadratic is

x^{2} - (α + β) x + α β = 0

\Rightarrow x^{2} - x - 6 = 0

Q. If the sum of the roots of the quadratic equations is $1$ and sum of the squares of the roots is $13$ , then find that equation

$x^{2} + x - 6 = 0$

$x^{2} - x + 6 = 0$

$x^{2} - x - 6 = 0$

$x^{2} + x + 6 = 0$

Q. If the sum of the roots of the quadratic equations is 1 and sum of the squares of the roots is 13, then find that equation

x2+x−6=0

x2−x+6=0

x2−x−6=0

x2+x+6=0

Let the roots of the quadratic equation are α and β then it is given that α+β=1 and α2+β2=13 ∴αβ=21​[(α+β)2−(α2+β2)] =21​[1−13]=−6 So, equation of required quadratic is x2−(α+β)x+αβ=0 ⇒x2−x−6=0

Q. If the sum of the roots of the quadratic equations is $1$ and sum of the squares of the roots is $13$ , then find that equation

$x^{2} + x - 6 = 0$

$x^{2} - x + 6 = 0$

$x^{2} - x - 6 = 0$

$x^{2} + x + 6 = 0$