Sum of roots is -1 and sum of their reciproca | vedclass

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

Question

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.Given that the sum of the roots is $\alpha + \beta = -1$.Given that the sum of the

Vedclass · Accepted Answer

$x^2 + x - 6 = 0$

Vedclass · Answer

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.Given that the sum of the roots is $\alpha + \beta = -1$.Given that the sum of their reciprocals is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{6}$.Simplifying the sum of reciprocals: $\frac{\alpha + \beta}{\alpha \beta} = \frac{1}{6}$.Substituting $\alpha + \beta = -1$ into the equation: $\frac{-1}{\alpha \beta} = \frac{1}{6}$,which gives $\alpha \beta = -6$.The standard form of a quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values: $x^2 - (-1)x + (-6) = 0$.Therefore,the equation is $x^2 + x - 6 = 0$.

Sum of roots is $-1$ and sum of their reciprocals is $\frac{1}{6}$,then the equation is

Similar Questions

If the roots of the equation $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$ are equal but opposite in sign,then the value of $m$ is:

Let $\alpha$ and $\beta$ be the roots of the equation $px^2 + qx + r = 0$ (where $p \neq 0$). If $p, q, r$ are in $A.P.$ and $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,then the value of $|\alpha - \beta|$ is

The quadratic equation whose one root is $\frac{1}{2 + \sqrt{5}}$ will be

Solution of the equation $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1$ is

The values of $a$ and $b$ for which the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module