Imaginary number comletiong square ?

Question

cant you also use the quadratic formula

Answer 1

>>question was find roots ix^2-3x-5i=0 solveequation

Dividing through by i gives you

x^2 - (3/i)x - 5 = 0

To complete the square, move the 5 over to the right, then take half of (3/i), square it, and add it to both sides.

x^2 - (3/i)x = 5

x^2 - (3/i)x + (3/2i)^2= 5 + (3/2i)^2

(x - (3/2)i)^2 = 5 + (9/ -4)

(x - (3/2)i)^2 = 11/4

x - (3/2)i = ±(√11)/2

x= (3i ± √11)/2

>>o yeah i meant why multiply by -i why not multiply

>>everything by i

Because that would give you -x^2 for the first term.  When completing the square, you want to start with the x^2 coefficient being 1.