If frac{alpha}{alpha+1} and frac{beta}{beta+1 | vedclass

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

Question

(A) Let $y = \frac{x}{x+1}$. Then $y(x+1) = x$ $\Rightarrow yx + y = x$ $\Rightarrow x(y-1) = -y$ $\Rightarrow x = \frac{y}{1-y}$.Since $\frac{\alpha}

Vedclass · Accepted Answer

$3x^2-x-3=0$

Vedclass · Answer

(A) Let $y = \frac{x}{x+1}$. Then $y(x+1) = x$ $\Rightarrow yx + y = x$ $\Rightarrow x(y-1) = -y$ $\Rightarrow x = \frac{y}{1-y}$.Since $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ are roots of $x^2+7x+3=0$,substituting $x = \frac{y}{1-y}$ into the equation gives the equation for $y$ (which are $\alpha$ and $\beta$):$(\frac{y}{1-y})^2 + 7(\frac{y}{1-y}) + 3 = 0$Multiply by $(1-y)^2$:$y^2 + 7y(1-y) + 3(1-y)^2 = 0$$y^2 + 7y - 7y^2 + 3(1 - 2y + y^2) = 0$$y^2 + 7y - 7y^2 + 3 - 6y + 3y^2 = 0$$(1-7+3)y^2 + (7-6)y + 3 = 0$$-3y^2 + y + 3 = 0$$3y^2 - y - 3 = 0$.Thus,the equation with roots $\alpha$ and $\beta$ is $3x^2-x-3=0$.

If $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ are the roots of the quadratic equation $x^2+7x+3=0$,then the equation having roots $\alpha$ and $\beta$ is

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?

For the equation $ax^2 + bx + c = 0$,the roots are $\alpha, \beta$,and for the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k, \beta - k$. Then $\frac{B^2 - 4AC}{b^2 - 4ac} = \dots$

If $1, 2, 3$ and $4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$,then $a+2b+c$ is equal to

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

If $\alpha, \beta$ are the roots of the equation $x^2 - 6x - 2 = 0$,$\alpha > \beta$ and $a_n = \alpha^n - \beta^n$,$n > 1$,then the value of $\frac{a_{10} - 2a_8}{2a_9}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module