Question

If $\dfrac{1}{{4 - 3i}}$ is a root of $a{x^2} + bx + 1 = 0,$ where $a,b$ are real, then
A.$a = 25\;,\;b = - 8$
B.$a = 25,\;b = 8$
C.$a = 5\;,\;b = 4$
D.None of these

Answer 1

Hint: The root given provided here is in complex form the quadratic equation have 2 roots are in the form of negative and positive imaginary no, like if $a + ib$ is one root, another root will be $a - ib.$

Complete step-by-step answer:
The given root of the equation $a{x^2} + bx + 1 = 0$ is $\dfrac{1}{{4 - 3i}}$ as we can see, the form of the root is complex but, complex form being in denominator doesn’t help to find the roots directly we need the complex form on the numerator part we can rationalize it to do do,
By rationalizing first root with $4 + 3i,$ we get
$ \Rightarrow \dfrac{1}{{4 - 3i}} \times \dfrac{{4 + 3i}}{{4 + 3i}}\; = \;\dfrac{{4 + 3i}}{{{4^2} - {{\left( {3i} \right)}^2}}} = \dfrac{{4 + 3i}}{{25}}$
As we know that the complex roots have complementary imaginary values. Which means that if $a + ib$ is one root then $a - bi$ is the other root.
If we follow the same we get the 2nd root as $\dfrac{{4 - 3i}}{{25}}$
If we have two roots ${z_1}$ and ${z_2}$ for equation of x, we can get equation by $\left( {x - {z_1}} \right),\left( {x - {z_2}} \right) = 0.$ similarly we can get equation for this by forming.
\[ \Rightarrow \left( {x - \dfrac{{4 + 3i}}{{25}}} \right)\left( {x - \dfrac{{4 - 3i}}{{25}}} \right) = 0\]
 $ \Rightarrow {x^2} - \left[ {\left( {\dfrac{{4 + 3i}}{{25}}} \right) + \left( {\dfrac{{4 - 3i}}{{25}}} \right)} \right]$ $x + \left( {\dfrac{{4 + 3i}}{{25}}} \right)\left( {\dfrac{{4 - 3i}}{{25}}} \right) = 0$
${x^2} - \left( {\dfrac{8}{{25}}} \right)x + \;\dfrac{{16 + 9}}{{25 \times 25}} = 0$
By taking LCM of 25, we get equation as $ \Rightarrow 25{x^2} - 8x + 1 = 0$
If we compare obtained equation with $a{x^2} + bx + 1 = 0$ we get $a = 25\,\,\,{\text{and}}\;\;{\text{b}}\;{\text{ = }}\; - 8$

Hence, $\left( {a,\;b} \right)\; = \left( {25, - 8} \right)$ option A is the correct Answer.

Note: Complex numbers also follows commutative law of Addition as well Multiplication
     $ * \;{z_1} + {z_2}\;\;\; = \;\;{z_2} + {z_1}$
     $ * \;{z_1} \times {z_2}\;\; = \;{z_2} \times {z_1}$
Where ${z_1}$ and ${z_2}$ are two different complex no.