Question

If one root of the equation $4{x^2} - 6x + p = 0{\text{ is }}q + 2i,$ where $p,q \in R$ then $p + q = 0$ ;
$\left( 1 \right)10$
$\left( 2 \right)19$
$\left( 3 \right) - 24$
$\left( 4 \right) - 32$

Answer 1

Hint: We should be familiar with the concept of quadratic equations in order to solve this question. The standard form of a quadratic equation is , $a{x^2} + bx + c = 0$ , where $a,{\text{ }}b,{\text{ }}c$ are coefficients of the equation and $a,b,c \in R$ . The degree (highest power of a variable) of a quadratic equation is $2$ and the degree of a quadratic equation gives us the number of roots it has, which means the number of roots for a quadratic equation will be $2$. There also exists a relationship between the coefficients and the roots of a quadratic equation and these formulae must be kept in mind while solving this question.

Complete step by step answer:
$ \Rightarrow 4{x^2} - 6x + p = 0{\text{ }}......\left( 1 \right)$
This is a quadratic equation ( An equation of degree $2$) that is why it has two roots or two zeroes. Let $\alpha {\text{ and }}\beta $ are the two roots of the given equation.
We have the following formulae for calculating Sum of roots and Product of roots;
$\left( 1 \right){\text{Sum of roots = }} - \left( {\dfrac{{{\text{Coefficient of }}x{\text{ }}}}{{{\text{Coefficient of }}{x^2}}}} \right)$
 $ \Rightarrow {\text{Sum of roots}} = \alpha + \beta = - \dfrac{b}{a}$
From equation $\left( 1 \right)$, we can say that ;
Coefficient of $x{\text{ is }} - 6$ and Coefficient of ${x^2}{\text{ is }}4$ , putting these values in the Sum of roots formula;
$ \Rightarrow {\text{Sum of roots}} = - \left( { - \dfrac{6}{4}} \right)$
$ \Rightarrow {\text{Sum of roots}} = \alpha + \beta = \dfrac{3}{2}{\text{ }}......\left( 2 \right)$
$\left( 2 \right){\text{Product of roots = }}\left( {\dfrac{{{\text{Constant term }}}}{{{\text{Coefficient of }}{x^2}}}} \right)$
From equation $\left( 1 \right)$, we can say that ;
Constant term i.e. $c = p$;
$ \Rightarrow {\text{Product of roots}} = {\text{ }}\dfrac{p}{4}{\text{ }}$
$ \Rightarrow {\text{Product of roots}} = \alpha \beta = \dfrac{p}{4}{\text{ }}......\left( 3 \right)$
Now from equation $\left( 2 \right)$ and equation $\left( 3 \right)$, we have the values;
$ \Rightarrow \alpha + \beta = \dfrac{3}{2}{\text{ and }}\alpha \beta {\text{ = }}\dfrac{p}{4}$
According to the question, one of the roots let say $\alpha = q + 2i$ ;
And let us suppose that $\beta = q - 2i$, then;
$ \Rightarrow \alpha + \beta = \left( {q + 2i} \right) + \left( {q - 2i} \right)$
That gives , $\alpha + \beta = 2q$;
From equation $\left( 2 \right)$, we know that $\alpha + \beta = \dfrac{3}{2}$
$ \Rightarrow \dfrac{3}{2} = 2q$
Therefore, we get the value of $q$ as;
$ \Rightarrow q = \dfrac{3}{4}{\text{ }}......\left( 4 \right)$
And then $ \Rightarrow \alpha \beta = \left( {q + 2i} \right)\left( {q - 2i} \right)$
$ \Rightarrow \alpha \beta = {q^2} - 2qi + 2qi - 4{i^2}$
We know the value of ${i^2} = - 1$
$ \Rightarrow \alpha \beta = {q^2} + 4$
From equation $\left( 3 \right)$, we know the value of $\alpha \beta = \dfrac{p}{4}$;
$ \Rightarrow \dfrac{p}{4} = {q^2} + 4$
Replace the value of $q = \dfrac{3}{4}{\text{ from equation }}\left( 4 \right),{\text{we get;}}$
$ \Rightarrow \dfrac{p}{4} = {\left( {\dfrac{3}{4}} \right)^2} + 4$
On further simplification;
$ \Rightarrow \dfrac{p}{4} = \dfrac{9}{{16}} + 4$
Solving the above equation to get the value of $p$ ;
$ \Rightarrow \dfrac{p}{4} = \dfrac{{73}}{{16}}$
Therefore, the value of $p = \dfrac{{73}}{4}{\text{ }}......\left( 5 \right)$
From ${\text{equation }}\left( 4 \right){\text{ and equation }}\left( 5 \right),{\text{ we have the values of }}q{\text{ and }}p{\text{ respectively;}}$
As asked in the question; the value of $p + q$ will be;
$ \Rightarrow p + q = \dfrac{{73}}{4} + \dfrac{3}{4}$
$ \Rightarrow p + q = \dfrac{{76}}{4}$
$ \Rightarrow p + q = 19$

So, the correct answer is “Option 2”.

Note: The roots of a quadratic equation are also called zeroes, because if we substitute the value of
roots in the equation then it becomes equal to zero. The roots are calculated by the standard quadratic
equation formula, i.e. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . The value under the radical sign i.e.
${b^2} - 4ac$ is called the discriminant and it is represented by ${\text{D}}$ . In this question, we can
notice one thing that the roots are imaginary and complex conjugate of each other means their real
parts are equal but imaginary parts differ in sign. This scenario occurs when the value of the discriminant
is negative i.e. ${b^2} - 4ac \prec 0$ .