Let alpha, beta be the roots of the equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\alpha + \beta = \sqrt{2}$ and $\alpha \beta = \sqrt{6}$.Let the roots of $x^{2}+ax+b=0$ be $y_1 = \frac{1}{\alpha^{2}}+1$ and $y_2 = \frac

Vedclass · Accepted Answer

real and both negative

Vedclass · Answer

(B) Given $\alpha + \beta = \sqrt{2}$ and $\alpha \beta = \sqrt{6}$.Let the roots of $x^{2}+ax+b=0$ be $y_1 = \frac{1}{\alpha^{2}}+1$ and $y_2 = \frac{1}{\beta^{2}}+1$.Sum of roots: $-a = y_1 + y_2 = \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} + 2 = \frac{(\alpha+\beta)^{2}-2\alpha \beta}{(\alpha \beta)^{2}} + 2 = \frac{2-2\sqrt{6}}{6} + 2 = \frac{1-\sqrt{6}}{3} + 2 = \frac{7-\sqrt{6}}{3}$.Product of roots: $b = y_1 y_2 = (\frac{1}{\alpha^{2}}+1)(\frac{1}{\beta^{2}}+1) = \frac{1}{(\alpha \beta)^{2}} + \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} + 1 = \frac{1}{6} + \frac{2-2\sqrt{6}}{6} + 1 = \frac{1+2-2\sqrt{6}+6}{6} = \frac{9-2\sqrt{6}}{6}$.Now,consider the equation $x^{2}-(a+b-2)x+(a+b+2)=0$.$a+b = -\frac{7-\sqrt{6}}{3} + \frac{9-2\sqrt{6}}{6} = \frac{-14+2\sqrt{6}+9-2\sqrt{6}}{6} = -\frac{5}{6}$.The equation becomes $x^{2}-(-\frac{5}{6}-2)x+(-\frac{5}{6}+2) = 0$,which is $x^{2} + \frac{17}{6}x + \frac{7}{6} = 0$,or $6x^{2}+17x+7=0$.Roots are $x = \frac{-17 \pm \sqrt{289 - 168}}{12} = \frac{-17 \pm \sqrt{121}}{12} = \frac{-17 \pm 11}{12}$.$x_1 = -\frac{28}{12} = -\frac{7}{3}$ and $x_2 = -\frac{6}{12} = -\frac{1}{2}$.Both roots are real and negative.

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 $x^{2}+ax+b=0$. Then the roots of the equation $x^{2}-(a+b-2)x+(a+b+2)=0$ are...

Similar Questions

If $S = {m \in \mathbb{R} : x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0}$ has distinct roots $\}$,then the number of elements in $S$ is

The least positive value of $a$ for which the equation $2x^{2} + (a-10)x + \frac{33}{2} = 2a$ has real roots is

The roots of $|x - 2|^2 + |x - 2| - 6 = 0$ are

The polynomial equation of degree $5$ whose roots are the roots of the equation $x^5-3x^4-x^3+11x^2-12x+4=0$ each increased by $2$,is

What is the remainder of $2x^3 - 5x^2 + 7$ when divided by $(x - 2)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module