Question

If one root of the equation $x^2+px+q=0$ is $2+\sqrt{3}$, then the values of $p\text{and}q$ are respectively

• A. $-4,1$
• B. $4,-1$
• C. $2,\sqrt{3}$
• D. $-2,-\sqrt{3}$

Answer 1

The correct option is A

$-4,1$

Explanation for the correct option

We know that if one of the roots of the quadratic equation is given then there will be a conjugate root for the same.

So, for the root $2+\sqrt{3}$ the conjugate base will be $2-\sqrt{3}$.

Formulae for the sum of the roots and product of the roots,

$$\begin{gathered} \mathbf{\text{Sum of the roots}}\mathbf{=}\mathbf{-}\frac{\mathbf{\text{Coefficient of}}\mathbf{}\mathbf{x}}{\mathbf{\text{Coefficient of}}\mathbf{}{\mathbf{x}}^{\mathbf{2}}} \\ \mathbf{\text{Product of the roots}}\mathbf{=}\frac{\mathbf{\text{Constant}}}{\mathbf{\text{Coefficient of}}\mathbf{}{\mathbf{x}}^{\mathbf{2}}} \end{gathered}$$

From the first formula, we have

$\begin{array}{rcl}2+\sqrt{3}+2-\sqrt{3}& =& -\frac{p}{1}\\ & \Rightarrow & p=-4\end{array}$

From the second formula, we have

$\begin{array}{rcl}\therefore q& =& \left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\\ & =& {\left(2\right)}^2-{\left(\sqrt{3}\right)}^2\\ & =& 1\end{array}$

Hence, the correct option is (A).