Let p, q in mathbb{Q}. If 2 - sqrt{3} is a ro | vedclass

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

Question

(B) Given that $p, q \in \mathbb{Q}$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.Since the coefficients are rational,the

Vedclass · Accepted Answer

$p^2 - 4q - 12 = 0$

Vedclass · Answer

(B) Given that $p, q \in \mathbb{Q}$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.Since the coefficients are rational,the irrational roots must occur in conjugate pairs.Therefore,the other root is $2 + \sqrt{3}$.Sum of roots $= (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.From the equation $x^2 + px + q = 0$,the sum of roots $= -p$.So,$-p = 4 \implies p = -4$.Product of roots $= (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.From the equation,the product of roots $= q$.So,$q = 1$.Now,check the options:For option $(B)$,$p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$.Thus,option $(B)$ is correct.

Let $p, q \in \mathbb{Q}$. If $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$,then:

Similar Questions

The set of values of $x \in R$ satisfying the inequality $x^2 - 4x - 21 \leq 0$ is

If $x = \sqrt[3]{{\sqrt 2 + 1}} - \sqrt[3]{{\sqrt 2 - 1}}$,then find the value of ${x^3} + 3x$.

Solve the equation $\sqrt{2} x^{2} + x + \sqrt{2} = 0$.

If $1-i$ is a root of the equation $x^2+ax+b=0$ where $a$ and $b$ are real numbers,then $b$ is equal to

If the equation $x^2 + \lambda x + \mu = 0$ has equal roots and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module