Let \(\alpha, \beta\) be the roots of the equation \(x^{2}-\sqrt{2} x+\sqrt{6}=0\) and \(\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1\) be the roots of the equation \(x^{2}+a x+b=0\). Then the roots of the equation \(\mathrm{x}^{2}-(\mathrm{a}+\mathrm{b}-2) \mathrm{x}+(\mathrm{a}+\mathrm{b}+2) =0\) are :

(D) real and exactly one of them is positive

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Quadratic Equations and Inequalities

Let

α, β

be the roots of the equation

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

and

\frac{1}{α ^{2}} + 1, \frac{1}{β ^{2}} + 1

be the roots of the equation

x^{2} + a x + b = 0

. Then the roots of the equation

x^{2} - (a + b - 2) x + (a + b + 2) = 0

are :

Answer: (B)

From

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

, by Vieta's formulas:

α + β = 2

α β = 6

.

For the second equation with roots

\frac{1}{α ^{2}} + 1

and

\frac{1}{β ^{2}} + 1

a = - (\frac{1}{α ^{2}} + \frac{1}{β ^{2}} + 2)

b = (\frac{1}{α ^{2}} + 1) (\frac{1}{β ^{2}} + 1) = \frac{1}{α ^{2} β ^{2}} + \frac{1}{α ^{2}} + \frac{1}{β ^{2}} + 1

Now

\frac{1}{α ^{2}} + \frac{1}{β ^{2}} = \frac{( α + β ) ^{2} - 2 α β}{( α β ) ^{2}} = \frac{2 - 2 6}{6} = \frac{1 - 6}{3}

And

\frac{1}{α ^{2} β ^{2}} = \frac{1}{6}

Thus

a + b = \frac{1}{6} - 1 = - \frac{5}{6}

The equation becomes:

x^{2} + \frac{17}{6} x + \frac{7}{6} = 0

6 x^{2} + 17 x + 7 = 0

Factoring:

(3 x + 1) (2 x + 7) = 0

, giving

x = - \frac{1}{3}, - \frac{7}{2}

. Both roots are real and negative.

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option