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

If $\alpha$ and $\beta$ are the real roots of the equation $\sqrt{\frac{5x}{x-2}} + \sqrt{\frac{x-2}{5x}} = \frac{29}{10}$ and $\alpha > \beta$,then $\sqrt{\alpha^2 - 11^4 \beta^2} = $

For what value of $k$ will the equation $x^2 - (3k - 1)x + 2k^2 + 2k = 0$ have equal roots?

If $ax^2 + bx + c = 0$,then $x =$

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

In $\triangle ABC$,the value of $\angle A$ is obtained from the equation $3 \cos A + 2 = 0$. The quadratic equation,whose roots are $\sin A$ and $\tan A$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module