If alpha and beta are the roots of x^2+7x+3=0 | vedclass

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

Question

(D) Let $y = \frac{2\alpha}{3-4\alpha}$. Then $2\alpha = 3y - 4\alpha y$,which implies $\alpha(2+4y) = 3y$,so $\alpha = \frac{3y}{2+4y}$.Since $\alpha

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(D) Let $y = \frac{2\alpha}{3-4\alpha}$. Then $2\alpha = 3y - 4\alpha y$,which implies $\alpha(2+4y) = 3y$,so $\alpha = \frac{3y}{2+4y}$.Since $\alpha$ is a root of $x^2+7x+3=0$,we substitute $\alpha$:$(\frac{3y}{2+4y})^2 + 7(\frac{3y}{2+4y}) + 3 = 0$.Multiplying by $(2+4y)^2 = 16y^2+16y+4$:$9y^2 + 21y(2+4y) + 3(16y^2+16y+4) = 0$.$9y^2 + 42y + 84y^2 + 48y^2 + 48y + 12 = 0$.$141y^2 + 90y + 12 = 0$.Dividing by $3$: $47y^2 + 30y + 4 = 0$.Comparing with $ax^2+bx+c=0$,we get $a=47, b=30, c=4$. Since $GCD(47, 30, 4) = 1$,the values are valid.Thus,$a+b+c = 47+30+4 = 81$.

If $\alpha$ and $\beta$ are the roots of $x^2+7x+3=0$ and $\frac{2\alpha}{3-4\alpha}, \frac{2\beta}{3-4\beta}$ are the roots of $ax^2+bx+c=0$ and $GCD(a, b, c) = 1$,then $a+b+c=$

Similar Questions

If the $A.M.$ and $G.M.$ of the roots of a quadratic equation are $8$ and $5$ respectively,then the quadratic equation 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 the roots of the equation $Ax^2 + Bx + C = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2, \beta^2$,then the value of $p$ is:

If $\alpha, \beta$ are the roots of the equation $x^2+3x+k=0$ and $\alpha+\frac{1}{\alpha}, \beta+\frac{1}{\beta}$ are the roots of the equation $4x^2+px+18=0$,then $k$ satisfies the equation:

The value of $p$ for which the sum of the squares of the roots of the equation $x^2 - (p + 3)x + (5p - 2) = 0$ assumes its least value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module