Question

If one root of the quadratic equation, \[i{x^2} - 2(i + 1)x + (2 - i) = 0\] is $2 - i$, then the other root is
A. $ - i$
B. $i$
C. $2 + i$
D. $2 - i$

Answer 1

Hint: In this question, we are given a quadratic equation \[i{x^2} - 2(i + 1)x + (2 - i) = 0\]. Also, one root of the equation is $2 - i$ and we have to find the other root. Let, the unknown root be $\alpha $. Then, apply the formula of product of roots i.e., $\alpha \beta = \dfrac{C}{A}$ whose quadratic equation is \[A{x^2} + Bx + C = 0\].

Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Given that,
\[i{x^2} - 2(i + 1)x + (2 - i) = 0\] , a quadratic equation
One root of the above equation $\beta = 2 - i$
Let, the other root of the equation be $\alpha $
Compare the given equation \[i{x^2} - 2(i + 1)x + (2 - i) = 0\] with the general quadratic equation i.e., \[A{x^2} + Bx + C = 0\]
It implies that,
$A = i$, $B = - 2\left( {i + 1} \right)$, and $C = \left( {2 - i} \right)$
As we know,
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Substitute the values in the above formula
We get,
$\alpha \left( {2 - i} \right) = \dfrac{{\left( {2 - i} \right)}}{i}$
Cancel the like terms of both the sides,
It implies that, $\alpha = \dfrac{1}{i}$
Multiply the numerator and denominator of the right side by $i$,
$\alpha = \dfrac{{1 \times i}}{{i \times i}}$
$\alpha = \dfrac{i}{{{i^2}}}$
As we know, the square of $i$ is equal to $ - 1$
Therefore, $\alpha = - i$
Thus, the roots of the equation \[i{x^2} - 2(i + 1)x + (2 - i) = 0\] are $ - i$ and $2 - i$.

Hence, the correct option is (A).

Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.