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

If $\alpha, \beta, \gamma$ are the real roots of the equation $x^3 - 3px^2 + 3qx - 1 = 0$,then find the centroid of the triangle whose vertices are $(\alpha, \frac{1}{\alpha}), (\beta, \frac{1}{\beta})$ and $(\gamma, \frac{1}{\gamma})$.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$ and $3b^{2}=16ac$, then:

If the $A.M.$ and $G.M.$ of the roots of a quadratic equation are $8$ and $5$ respectively,then the quadratic equation is:

Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2 - px + q = 0$,then the value of $\sin^2(\alpha + \beta)$ is:

For the equation $ax^2 + bx + c = 0$,if $\alpha$ and $\beta$ are its roots,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module