Question

The quadratic equation with one root $\dfrac{1}{2}\left( 1+\sqrt{-3} \right)$ is –
A. ${{x}^{2}}-x-1=0$
B ${{x}^{2}}+x-1=0$
C. ${{x}^{2}}+x+1=0$
D. ${{x}^{2}}-x+1=0$

Answer 1

Hint: To solve this question, we will make use of the fact that if any equation whose coefficient of ${{x}^{2}}$ The coefficients of x and constant term all are real. Then the roots of that equation will be in conjugate form if the roots are irrational or complex in nature. Thus, if one root of the quadratic equation is $a+ib$ then the other root of that quadratic equation will be $a-ib$ .

Complete step-by-step answer:
Before we solve the question given we must know what is a quadratic equation. A quadratic equation in x is the equation which contains maximum power of x equal to two. Thus, the general form of any quadratic equation will be $a{{x}^{2}}+bx+c=0$ . the other form of quadratic equation is $\left( x-\alpha \right)\left( x-\beta \right)=0$ , where $\alpha $ and $\beta $ are the roots of the given quadratic equation. Now, we are given that one root of a quadratic equation is $x=\dfrac{1}{2}\left( 1+\sqrt{-3} \right)$ and we have to find another root of this equation. For this, we will assume that the coefficient of ${{x}^{2}}$ , coefficient of x and constant term, all will be real. We have made such assumptions because of options. We can see that all the quadratic equations given in the options have real coefficients. Now, we will make use of the fact that if coefficients of a quadratic equation are real and one of the roots is complex in nature then the other root will be the conjugate of the first one. Thus, if a quadratic equation has one root equal to $a+ib$ then the other root will bet $a-ib$ . we are given that one root of a quadratic equation is $\dfrac{1}{2}\left( 1+\sqrt{-3} \right)$ . Thus, we have:
$\begin{align}
  & \alpha =\dfrac{1}{2}\left( 1+\sqrt{-3} \right) \\
 & \Rightarrow \alpha =\dfrac{1}{2}\left( 1+\sqrt{3}i \right) \\
 & \Rightarrow \alpha =\dfrac{1}{2}+\dfrac{\sqrt{3}i}{2} \\
\end{align}$
So, the other root $\beta =\dfrac{1}{2}-\dfrac{\sqrt{3}i}{2}$
Now, we know that a quadratic equation can be written as:
$\left( x-\alpha \right)\left( x-\beta \right)=0$ .
Thus we have:
$\begin{align}
  & \left[ x-\left( \dfrac{1}{2}-\dfrac{\sqrt{3}i}{2} \right) \right]\left[ x-\left( \dfrac{1}{2}+\dfrac{\sqrt{3}i}{2} \right) \right]=0 \\
 & \Rightarrow \left( x-\dfrac{1}{2}+\dfrac{\sqrt{3}i}{2} \right)\left( x-\dfrac{1}{2}-\dfrac{\sqrt{3}i}{2} \right)=0 \\
 & \Rightarrow \left[ \left( x-\dfrac{1}{2} \right)+\left( \dfrac{\sqrt{3}i}{2} \right) \right]\left[ \left( x-\dfrac{1}{2} \right)-\left( \dfrac{\sqrt{3}i}{2} \right) \right]=0 \\
\end{align}$
Here, we are going to use the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ . Thus, we have:
${{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}i}{2} \right)}^{2}}=0$
Now, we will use the identity ${{\left( a-b \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}-2ab \right)$
$\begin{align}
  & {{x}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{1}{2} \right)x-\dfrac{3{{\left( i \right)}^{2}}}{4}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{4}-x+\dfrac{3\left( -1 \right)}{4}=0\text{ }\left\{ {{i}^{2}}=-1 \right\} \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{4}-x+\dfrac{3}{4}=0 \\
 & \Rightarrow {{x}^{2}}-x+1=0 \\
\end{align}$
Hence, option (d) is correct.

Note: In this question, it is important to assume that the coefficient quadratic equations are real numbers. If we do not consider this, then we will not be able to say that the roots are conjugate in nature if one of the roots is complex. If the coefficients of the quadratic equation are complex, then the roots would not be conjugate in nature.