If 2 and 6 are the roots of the equation ax^2 | vedclass

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

Question

(D) Given the roots of $ax^2 + bx + 1 = 0$ are $2$ and $6$. Sum of roots: $2 + 6 = 8 = -\frac{b}{a} \implies b = -8a$. Product of roots: $2 	imes 6 =

Vedclass · Accepted Answer

$x^2 + 8x + 12 = 0$

Vedclass · Answer

(D) Given the roots of $ax^2 + bx + 1 = 0$ are $2$ and $6$. Sum of roots: $2 + 6 = 8 = -\frac{b}{a} \implies b = -8a$. Product of roots: $2 	imes 6 = 12 = \frac{1}{a} \implies a = \frac{1}{12}$. Substituting $a$ into $b = -8a$: $b = -8 	imes \frac{1}{12} = -\frac{2}{3}$. Now,calculate the new roots: Root $1 = \frac{1}{2a + b} = \frac{1}{2(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{6} - \frac{4}{6}} = \frac{1}{-\frac{3}{6}} = -2$. Root $2 = \frac{1}{6a + b} = \frac{1}{6(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{2} - \frac{2}{3}} = \frac{1}{-\frac{1}{6}} = -6$. The quadratic equation with roots $-2$ and $-6$ is $(x + 2)(x + 6) = 0$. $x^2 + 8x + 12 = 0$.

If $2$ and $6$ are the roots of the equation $ax^2 + bx + 1 = 0$,then the quadratic equation,whose roots are $\frac{1}{2a + b}$ and $\frac{1}{6a + b}$,is:

Similar Questions

Assume that $\alpha, \beta, \gamma$ are the roots of $2x^3+5x^2+5x+2=0$. For $h \in R$,if $\alpha+h, \beta+h, \gamma+h$ are roots of $a(h)x^3+b(h)x^2+c(h)x+d(h)=0$,then:

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

Let $\alpha$ and $\beta$ be the roots of $x^{2}-6x-2=0$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$,then the value of $\frac{a_{10}-2a_{8}}{3a_{9}}$ is ..... .

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - a(x + 1) - b = 0$,then $(\alpha + 1)(\beta + 1) = $

Let $\alpha, \beta$ be roots of $x^2+\sqrt{2}x-8=0$. If $U_n = \alpha^n + \beta^n$,then $\frac{U_{10} + \sqrt{2}U_9}{2U_8}$ is equal to ............

Vedclass Test Series

Exam Paper Generator

Online Exam Module