x^4+2ix-3x^2-yi=3-5i+1+2iy what are the values of x and y

Question

terenzreignz · Answer

First, simplify the right part.  Might as well do the left part, too...

anonymous · Answer

x^4-3x^2=4 and 2x-3y+5=0

anonymous · Answer

plz simplify furthur

terenzreignz · Answer

You can solve for x in this equation
$$\huge x^4-3x^2 = 4$$

terenzreignz · Answer

For which there are many possible values, possibly...
$$\huge x^4 - 3x^2 - 4 = 0$$

terenzreignz · Answer

Oh, and before you panic, maybe you can let $$\huge u = x^2$$and solve for u instead...

terenzreignz · Answer

$$\huge u^2-3u-4 = 0$$

terenzreignz · Answer

@leena1996 work with me here.

terenzreignz · Answer

I do hope x is only allowed to take real number values, though....

anonymous · Answer

ya i got x=+/-2 and y should be 1/3 but i am getting it as 3....

terenzreignz · Answer

I wonder why...
$$\huge 2x -3y + 5 = 0 $$$$\huge2(\pm2)-3y +5 = 0$$$$\huge \pm4-3y=-5$$$$\huge -3y=-5\mp4$$

$$\huge y = \frac{5\pm4}{3}$$

terenzreignz · Answer

It seems both 1/3 and 3 are possible, @leena1996 
depending on which value of x you pick ...
If x = 2, then y = 3
If x = -2, then y = 1/3