Solve: i x^2 - 3 x - 2 i = 0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $i x^2 - 3 x - 2 i = 0$Since $i^2 = -1$,we can write $-3x$ as $i^2 x - 4x$ or manipulate the middle term.Alternatively,using the q

Vedclass · Accepted Answer

$-i$ and $-2 i$

Vedclass · Answer

(C) Given equation: $i x^2 - 3 x - 2 i = 0$Since $i^2 = -1$,we can write $-3x$ as $i^2 x - 4x$ or manipulate the middle term.Alternatively,using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:Here $a = i$,$b = -3$,$c = -2i$.$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(i)(-2i)}}{2(i)}$$x = \frac{3 \pm \sqrt{9 + 8i^2}}{2i}$$x = \frac{3 \pm \sqrt{9 - 8}}{2i}$$x = \frac{3 \pm \sqrt{1}}{2i}$$x = \frac{3 \pm 1}{2i}$Case $1$: $x = \frac{3+1}{2i} = \frac{4}{2i} = \frac{2}{i} = -2i$Case $2$: $x = \frac{3-1}{2i} = \frac{2}{2i} = \frac{1}{i} = -i$Thus,the solutions are $x = -i$ and $x = -2i$.

Solve: $i x^2 - 3 x - 2 i = 0$

Similar Questions

If $z=x+iy$ and $z^{1/3}=p+iq$,where $x, y, p, q \in R$ and $i=\sqrt{-1}$,then the value of $\left(\frac{x}{p}+\frac{y}{q}\right)$ is

The complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ is equal to $.....$

If $z$ is a complex number satisfying $|z^3+z^{-3}| \leq 2$,then the maximum possible value of $|z+z^{-1}|$ is

Suppose $z$ is any root of $11 z^8 + 21 i z^7 + 10 i z - 22 = 0$ where $i = \sqrt{-1}$. Then,$S = |z|^2 + |z| + 1$ satisfies

Let $z=x+iy$ and $w=u+iv$ be complex numbers on the unit circle such that $z^2+w^2=1$. Then the number of ordered pairs $(z, w)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module