If the roots of the quadratic equation x^2 + | vedclass

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

Question

(D) Given the quadratic equation $x^2 + px + q = 0$ with roots $\alpha = 	an 30^\circ$ and $\beta = 	an 15^\circ$.From the properties of roots,the s

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given the quadratic equation $x^2 + px + q = 0$ with roots $\alpha = 	an 30^\circ$ and $\beta = 	an 15^\circ$.From the properties of roots,the sum of roots is $\alpha + \beta = -p$ and the product of roots is $\alpha \beta = q$.We know that $	an(30^\circ + 15^\circ) = 	an 45^\circ = 1$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$1 = \frac{-p}{1 - q}$.This implies $1 - q = -p$,or $p - q = -1$,which can be rewritten as $p + q = 1$ is incorrect; rather $1 - q = -p$ $\Rightarrow p - q = -1$ $\Rightarrow q - p = 1$.We need to find the value of $2 + q - p$.Substituting $q - p = 1$,we get $2 + (1) = 3$.

If the roots of the quadratic equation $x^2 + px + q = 0$ are $\tan 30^\circ$ and $\tan 15^\circ$ respectively,then the value of $2 + q - p$ is

Similar Questions

Let $P_n = \alpha^n + \beta^n, n \in N$. If $P_{10} = 123, P_9 = 76, P_8 = 47$ and $P_1 = 1$,then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

If the roots of the equation $x^3+3px^2+3qx-8=0$ are in a geometric progression,then $\frac{q^3}{p^3}=$

If $\alpha, \beta, \gamma$ are the roots of the equation $3x^3 - 9x^2 + 5x - 7 = 0$,then what is the value of $\alpha + \beta + \gamma$?

If the roots of the equation $x^2 + px + q = 0$ are $p$ and $q$,then what must be the value of $p$?

The harmonic mean of two numbers is $-\frac{8}{5}$ and their geometric mean is $2$. The quadratic equation whose roots are twice those numbers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module