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
\[\begin{align}
  & \left( \alpha +i\beta \right)+\left( \alpha -i\beta \right)=-a+\left( -\overline{a} \right) \\
 & 2\alpha =-\left( a+\overline{a} \right) \to (3) \\
\end{align}\]
If we subtract (1) and (2) we get,
\[\begin{align}
  & \left( \alpha +i\beta \right)-\left( \alpha -i\beta \right)=-a-\left( -\overline{a} \right) \\
 & 2i\beta =-a+\overline{a} \to (4) \\
\end{align}\]
Now let us take the product of (3) and (4).
\[\begin{align}
  & \left( 2\alpha \right)\left( 2i\beta \right)=\left[ -\left( a+\overline{a} \right) \right]\left[ -\left( a-\overline{a} \right) \right] \\
 & 4\alpha \beta i=\left( a+\overline{a} \right)\left( a-\overline{a} \right) \\
\end{align}\]
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 \[\alpha \beta i=b\].