Question

If $2 + i\sqrt 3$ is a root of the equation ${x^2} + px + q = 0$ where $p$ and $q$ are real, then $\left( {p,q} \right)$ is equal to
A) $\left( { - 4,7} \right)$
B) $\left( {4, - 7} \right)$
C) $\left( {4,7} \right)$
D) $\left( { - 4, - 7} \right)$

Answer 1

Here we will use the concept of the roots as root satisfies the equation. Then we will put the value of the root in the equation and we will solve the equation. Then we will take the constant terms in one bracket and imaginary in another bracket. Then we will equate these equations to zero to get the value of $p$ and $q$.

Complete Step by step Solution:
Given equation is ${x^2} + px + q = 0$ and the given root of the equation is $2 + i\sqrt 3$.
We know that the root of an equation satisfies the equation when it is put in the variable term. Therefore, we get
${\left( {2 + i\sqrt 3 } \right)^2} + p\left( {2 + i\sqrt 3 } \right) + q = 0$
Now we will use the basic algebraic identity to expand the square of the term i.e. ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. Therefore, we get
$\Rightarrow {\left( 2 \right)^2} + {\left( {i\sqrt 3 } \right)^2} + \left( {2 \times 2 \times i\sqrt 3 } \right) + p\left( {2 + i\sqrt 3 } \right) + q = 0$
Now we will simply solve this equation, we get
$\Rightarrow 4 + 3{i^2} + 4\sqrt 3 i + 2p + \sqrt 3 ip + q = 0$
We know that the value of ${i^2} = - 1$. Therefore, we get
$\Rightarrow 4 - 3 + 4\sqrt 3 i + 2p + \sqrt 3 ip + q = 0$
$\Rightarrow 1 + 4\sqrt 3 i + 2p + \sqrt 3 ip + q = 0$
Now we will take the constant terms in one bracket of the equation and take imaginary terms in the other bracket of the equation. Therefore, we get
$\Rightarrow \left( {1 + 2p + q} \right) + \left( {4\sqrt 3 + \sqrt 3 p} \right)i = 0$
Now from the above equation we get two equations. First equation is
$\Rightarrow \left( {1 + 2p + q} \right) = 0$
$\Rightarrow 2p + q = - 1$………………………..$\left( 1 \right)$
Second equation is
$\Rightarrow 4\sqrt 3 + \sqrt 3 p = 0$
By solving this second equation we will get the value of $p$. Therefore, we get
$\Rightarrow \sqrt 3 p = - 4\sqrt 3$
$\Rightarrow p = - 4$
Now we will put the value of $p$ in the equation $\left( 1 \right)$ to get the value of $q$. Therefore, we get
$\Rightarrow 2\left( { - 4} \right) + q = - 1$
$\Rightarrow - 8 + q = - 1$
$\Rightarrow q = - 1 + 8 = 7$
Hence the value of $\left( {p,q} \right)$ is $\left( { - 4,7} \right)$.

So, option A is the correct option.

Note:
Roots are those values of the equation where the value of the equation becomes zero. For any equation numbers of roots are always equal to the value of the highest exponent of the variable x. Linear equation is the equation in which the highest exponent of the variable x is one. A quadratic equation is an equation in which the highest exponent of the variable x is two and a quadratic equation has only two roots. Also, the cubic equations are the equation in which the highest exponent of the variable is 3. Also, we know that for an equation the number of its roots is equal to the value of the highest exponent of the variable of that equation which means for a cubic equation there are three roots of the equation.
Quadratic equation: $a{x^2} + bx + c = 0$
Cubic equation: $a{x^3} + b{x^2} + cx + d = 0$