If alpha, beta are the roots of the equation | vedclass

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

Question

(D) For the equation $x^2 - 5x + 16 = 0$,the sum of roots $\alpha + \beta = 5$ and the product of roots $\alpha\beta = 16$.We need to find the roots o

Vedclass · Accepted Answer

$p = -1, q = -56$

Vedclass · Answer

(D) For the equation $x^2 - 5x + 16 = 0$,the sum of roots $\alpha + \beta = 5$ and the product of roots $\alpha\beta = 16$.We need to find the roots of the second equation $x^2 + px + q = 0$.The first root is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (5)^2 - 2(16) = 25 - 32 = -7$.The second root is $\frac{\alpha\beta}{2} = \frac{16}{2} = 8$.The roots of $x^2 + px + q = 0$ are $-7$ and $8$.Sum of roots: $-7 + 8 = 1 = -p \implies p = -1$.Product of roots: $(-7)(8) = -56 = q \implies q = -56$.Thus,$p = -1$ and $q = -56$.

If $\alpha, \beta$ are the roots of the equation $x^2 - 5x + 16 = 0$ and $(\alpha^2 + \beta^2)$ and $\frac{\alpha\beta}{2}$ are the roots of the equation $x^2 + px + q = 0$,then:

Similar Questions

If the sum of the roots of the equation $x^2 + px + q = 0$ is three times their difference,then which one of the following is true?

If $\alpha$ and $\beta$ are two real numbers satisfying $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$,then the value of $\alpha \beta$ is:

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+ax^2+bx+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 - x - 1 = 0$,then find the equation whose roots are $\frac{1}{\beta + \gamma}, \frac{1}{\gamma + \alpha}, \frac{1}{\alpha + \beta}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module