If the difference between the roots of the eq | vedclass

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

Question

(C) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + ax + 1 = 0$.From the properties of quadratic equations,we have $\alpha + \beta = -a$

Vedclass · Accepted Answer

$ (-3, 3) $

Vedclass · Answer

(C) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + ax + 1 = 0$.From the properties of quadratic equations,we have $\alpha + \beta = -a$ and $\alpha \beta = 1$.The difference between the roots is given by $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.Substituting the values,we get $|\alpha - \beta| = \sqrt{(-a)^2 - 4(1)} = \sqrt{a^2 - 4}$.According to the given condition,$|\alpha - \beta| Therefore,$\sqrt{a^2 - 4} Squaring both sides,we get $a^2 - 4 This implies $|a| Thus,the set of possible values of $a$ is $a \in (-3, 3)$.

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

Similar Questions

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

The values of $a$ for which $2x^2 - 2(2a + 1)x + a(a + 1) = 0$ may have one root less than $a$ and other root greater than $a$ are given by

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} \times 13 = x^{2} \times 247$
$II.$ $y^{1 / 3} \times 14 = 294 \div y^{2 / 3}$

The value of $k$ for which the roots $\alpha, \beta$ of the equation $x^{2}-6x+k=0$ satisfy the relation $3\alpha+2\beta=20$ is:

Solve the given two equations and select the correct option.
$I.$ $\frac{x}{x-11} + \frac{x-11}{x} = 7$
$II.$ $\frac{4y}{4y-13} + \frac{4y-13}{4y} = 9$

Vedclass Test Series

Exam Paper Generator

Online Exam Module