Write in the form a+bi: (1-3i)/(-2-4i)?

Question

Not a kid with a home work problem, for a change... but an adult looking at Algebra II problem and wondering why he doesn't get their answer... obviously, I've forgotten how to do these. Show work please! Many thanks.

? · Accepted Answer

one way...

multiply numerator and denominator by conjugate. 

notes: for two terms added/subtracted, the conjugate is same two terms with opposite sign between.

so conjugate of -2 - 4i is -2 + 4i

also, as long as we multiply top & bottom of fraction by same (non-zero) value, the fraction still represents original value.

so...

(1 - 3i)/(-2 - 4i)

= [(1 - 3i)(-2 + 4i)] / [(-2 - 4i)(-2 + 4i)]

now expand top & bottom (with FOIL, diff.sqs., etc)

= (-2 + 4i + 6i - 12i²) / (4 - 8i + 8i - 16i²)

= (-2 + 10i - 12i²) / (4 - 16i²)

recall that i² = -1 (the complex #'s), so replace i² with -1 to get

= (-2 + 10i + 12) / (4 + 16)

= (-10 + 10i) / 20

= -10/20 + 10i/20

= -1/2 + (1/2)i

or

= -0.5 + 0.5i


hope that helps

Anonymous · Answer

(x - 3)^2 = 36, so (x - 3)(x - 3) = 36, so x(x - 3) - 3(x - 3) = 36, so x^2 - 3x - 3x + 9 = 36, so x^2 - 6x + 9 = 36, so x^2 - 6x - 27 = 0..........it is now in countless circumstances kind, ax^2 + bx + c = 0 Factoring innovations-set to sparkling up: x^2 - 6x - 27 = 0, so (x - 9)(x + 3) = 0, so x - 9 = 0, or x + 3 = 0, so x = 9, or x = -3.

Trending News

Write in the form a+bi: (1-3i)/(-2-4i)?

2 Answers