If the roots of x^2 - 7x + 6 = 0 are alpha an | vedclass

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

Question

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,the sum of the roots is $\alpha + \beta = -b/a$ and the product of the roots is $\alpha\b

Vedclass · Accepted Answer

$7/6$

Vedclass · Answer

(B) For a quadratic equation of the form $ax^2 + bx + c = 0$,the sum of the roots is $\alpha + \beta = -b/a$ and the product of the roots is $\alpha\beta = c/a$.Given the equation $x^2 - 7x + 6 = 0$,we have $a = 1$,$b = -7$,and $c = 6$.Therefore,the sum of the roots is $\alpha + \beta = -(-7)/1 = 7$.The product of the roots is $\alpha\beta = 6/1 = 6$.We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.By taking the common denominator,we get $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$.Substituting the values,we get $\frac{7}{6}$.

If the roots of $x^2 - 7x + 6 = 0$ are $\alpha$ and $\beta$,then $\frac{1}{\alpha} + \frac{1}{\beta} = $

Similar Questions

If the roots of the quadratic equation $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$ are reciprocal to each other,then

The solution of the equation $\frac{p + q - x}{r} + \frac{q + r - x}{p} + \frac{r + p - x}{q} + \frac{3x}{p + q + r} = 0$ is

If $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$,then

Solve the given two equations and choose the correct answer from the given options.
$I.$ $16x^2 + 20x + 6 = 0$
$II.$ $10y^2 + 38y + 24 = 0$

Solve the given two equations and select the correct option.
$I.$ $12x^2 - 47x + 40 = 0$
$II.$ $4y^2 + 3y - 10 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module