Question

Let $p,q\in R$ and if $2-\sqrt{3}$ is a root of the quadratic equation ${{x}^{2}}+px+q=0$, then which of the following options are true?
(a) ${{q}^{2}}+4p+14=0$,
(b) ${{p}^{2}}-4q-12=0$,
(c) ${{q}^{2}}-4p-16=0$,
(d) ${{p}^{2}}-4q+12=0$.

Answer 1

Hint: We start solving the problem by using the fact that if the quadratic equation has an irrational root, then the other root will be conjugate of the given irrational root. We then find the sum and product of the roots by using this fact. We then check options one after another to get the required answer.

Complete step-by-step answer:
According to the problem, $2-\sqrt{3}$ is a root of the quadratic equation ${{x}^{2}}+px+q=0$ $\forall p,q\in R$. We need to find which of the given options are true.
We know that if the quadratic equation has an irrational root, then the other root will be conjugate of the given irrational root i.e., if $r-\sqrt{s}$ is the root of the quadratic equation $a{{x}^{2}}+bx+c=0$, then $r+\sqrt{s}$ is also the root of that quadratic equation.
Since given root $2-\sqrt{3}$ is irrational and root of the quadratic equation ${{x}^{2}}+px+q=0$, then $2+\sqrt{3}$ is also root of the quadratic equation ${{x}^{2}}+px+q=0$.
We have $2-\sqrt{3}$ and $2+\sqrt{3}$ as roots of the quadratic equation ${{x}^{2}}+px+q=0$.
We know that sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$.
We have sum and product of roots of the quadratic equation ${{x}^{2}}+px+q=0$ as $-p$ and $q$.
We have $-p=2-\sqrt{3}+2+\sqrt{3}$.
$\Rightarrow -p=4$.
$\Rightarrow p=-4$ ---(1).
We have $q=\left( 2-\sqrt{3} \right)\times \left( 2+\sqrt{3} \right)$.
$\Rightarrow q=4-3$.
$\Rightarrow q=1$ ---(2).
Let us check for option (1).
We have ${{q}^{2}}+4p+14=0$.
$\Rightarrow {{1}^{2}}+4\left( -4 \right)+14=0$.
$\Rightarrow 1-16+14=0$.
$\Rightarrow -1=0$. Which is a contradiction.
Let us check for option (2).
We have ${{p}^{2}}-4q-12=0$.
$\Rightarrow {{\left( -4 \right)}^{2}}-4\left( 1 \right)-12=0$.
$\Rightarrow 16-4-12=0$.
$\Rightarrow 0=0$. Which is true.
Let us check for option (3).
We have ${{q}^{2}}-4p-16=0$.
$\Rightarrow {{1}^{2}}-4\left( -4 \right)-16=0$.
$\Rightarrow 1+16-16=0$.
$\Rightarrow 1=0$. Which is a contradiction.
Let us check for option (4).
We have ${{p}^{2}}-4q+12=0$.
$\Rightarrow {{\left( -4 \right)}^{2}}-4\left( 1 \right)+12=0$.
$\Rightarrow 16-4+12=0$.
$\Rightarrow 24=0$. Which is a contradiction.
We have found option (2) is correct.
The correct option for the given problem is (2).

Note: e should know conjugates of irrational and complex roots are also the roots of the given quadratic equation. We should not be confused about the sum and product of the roots of the quadratic equation. We should not make mistakes while doing addition, multiplication and signs of the roots. We can also expect to find roots if the sum and product of the roots are given.