If x^4+2xi-3x^2+yi=3-5i+1+2yi, then the number of ordered pairs x,y is/are equal to ∀x,y∈R and i^2=-1

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) 2. x^4-3x^2-3-1=0 & 2x-y+5-2y=0 x^2-4x^2+1=0 & 2x-3y+5=0 ⇒x=2 or -2 & 2x-3y+5=0 When x=2,y=3 and when x=-2,y=1/3 ⇒x,y=2,3 & -2,1/3 ⇒ Number of ordered pair of x,y is equal to 2

If $(x^{4} + 2 x i) - (3 x^{2} + y i) = (3 - 5 i) + (1 + 2 y i)$ , then the number of ordered pairs $(x, y)$ is/are equal to $\{\forall x, y \in R a n d i^{2} = - 1\}$

Solution

More Complex Number Questions

If x4+2xi-3x2+yi=3-5i+1+2yi, then the number of ordered pairs x,y is/are equal to ∀x,y∈R and i2=-1

Solution

More Complex Number Questions

If $(x^{4} + 2 x i) - (3 x^{2} + y i) = (3 - 5 i) + (1 + 2 y i)$ , then the number of ordered pairs $(x, y)$ is/are equal to $\{\forall x, y \in R a n d i^{2} = - 1\}$