If one root of the equation i x^2 - 2(i + 1) | vedclass

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Question

(A) Given the quadratic equation: $i x^2 - 2(i + 1) x + (2 - i) = 0$. Let the roots of the equation be $\alpha$ and $\beta$. We are given one root $\a

Vedclass · Accepted Answer

$-i$

Vedclass · Answer

(A) Given the quadratic equation: $i x^2 - 2(i + 1) x + (2 - i) = 0$. Let the roots of the equation be $\alpha$ and $\beta$. We are given one root $\alpha = 2 - i$. According to the relation between roots and coefficients,the product of the roots $\alpha 	imes \beta = \frac{c}{a}$. Here,$a = i$ and $c = 2 - i$. So,$(2 - i) 	imes \beta = \frac{2 - i}{i}$. Dividing both sides by $(2 - i)$,we get $\beta = \frac{1}{i}$. Multiplying the numerator and denominator by $i$,we get $\beta = \frac{i}{i^2} = \frac{i}{-1} = -i$. Thus,the other root is $-i$.

If one root of the equation $i x^2 - 2(i + 1) x + (2 - i) = 0$ is $(2 - i)$,then the other root is

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