Question

If a, b are complex number and one of the roots of the equation ${{x}^{2}}+ax+b=0$ is purely real, whereas the other is purely imaginary and ${{a}^{2}}-{{a}^{-2}}=kb$, then k is
(a) 2
(b) 4
(c) 6
(d) 8

Answer 1

Hint: Consider 2 roots as $\alpha$ and $i\beta$. Thus find $\left( \alpha +i\beta \right)$ which is the sum of zeroes of the root and its conjugate $\left( \alpha -i\beta \right)$. Now add and subtract these 2 equations, we get 2 equations. Now take their product and simplify it. Compare this equation with ${{a}^{2}}-{{a}^{-2}}=kb$ to get the value of k.

Complete step-by-step answer:
It is said that a and b are complex numbers. Then again we are given an equation, ${{x}^{2}}+ax+b=0$.
Now this equation has one pure real root and one imaginary root.
Now we need to find the value of k from ${{a}^{2}}-{{a}^{-2}}=kb$.
Let us consider $\alpha$ and $i\beta$ are the two roots of equation ${{x}^{2}}+ax+b=0$, where $\alpha$ is the real root and $i\beta$ is the pure imaginary root.
We know that a complex number is of the form $a+ib$. Thus the roots will form $\alpha +i\beta$. Thus,
$\alpha +i\beta =-a$, from the equation ${{x}^{2}}+ax+b$, which is the sum of zeroes.
$\therefore \alpha -i\beta =-\overline{a}$, where $\left( \alpha -i\beta \right)$ is the conjugate of $\left( \alpha +i\beta \right)$.
Thus we can say that, $\alpha +i\beta =-a \to (1)$
$\alpha -i\beta =-\overline{a} \to (2)$
Now let us add both these equations, we get

& \left( \alpha +i\beta \right)+\left( \alpha -i\beta \right)=-a+\left( -\overline{a} \right) \\\ & 2\alpha =-\left( a+\overline{a} \right) \to (3) \\\ \end{aligned}$$ If we subtract (1) and (2) we get, $$\begin{aligned} & \left( \alpha +i\beta \right)-\left( \alpha -i\beta \right)=-a-\left( -\overline{a} \right) \\\ & 2i\beta =-a+\overline{a} \to (4) \\\ \end{aligned}$$ Now let us take the product of (3) and (4). $$\begin{aligned} & \left( 2\alpha \right)\left( 2i\beta \right)=\left[ -\left( a+\overline{a} \right) \right]\left[ -\left( a-\overline{a} \right) \right] \\\ & 4\alpha \beta i=\left( a+\overline{a} \right)\left( a-\overline{a} \right) \\\ \end{aligned}$$ Now $$i\beta $$ is the product of 2 roots of equation $${{x}^{2}}+ax+b=0$$. Thus, $$i\alpha \beta =b$$, i.e. the product of zeroes. Hence we can write, $$4i\alpha \beta =\left( a+\overline{a} \right)\left( a-\overline{a} \right)$$ $$4b={{a}^{2}}-{{\left( \overline{a} \right)}^{2}}\left\\{ \because \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \right\\}$$ Now if we compare the equation, $${{a}^{2}}-{{\left( \overline{a} \right)}^{2}}=4b$$ and $${{a}^{2}}-{{\left( \overline{a} \right)}^{2}}=kb$$, we get k = 4. Thus we found the value of k = 4. $$\therefore $$ Option (b) is the correct answer. Note: A quadratic equation, $$a{{x}^{2}}+bx+c=0$$ is of the form as, $${{x}^{2}}$$ + (sum of zeroes) x + (product of zeroes) = 0. Here roots are $$\left( \alpha -i\beta \right)$$. Thus their sum, $$\alpha +i\beta =\dfrac{-a}{1}$$ = -coefficient of x / coefficient $${{x}^{2}}$$. Similarly, product $$\alpha \beta i=b$$.