Complex Number Equation (1 Viewer)

Utility · Aug 1, 2012

Solve (x²+6) + xi = 3(2+3i)(1-i)

I ended up with (x²+6) + xi = 15 + 3i & solved xi = 3i to get x = 3.

I'm wondering how else people do this?

SpiralFlex · Aug 1, 2012

$x^2+6+xi=3(2-2i+3i+3)$

$x^2+6+xi=15+3i$

$By equating imaginary coefficients,$

$x=3$

seanieg89 · Aug 1, 2012

Is it stated that x is real? If not then there is a second solution.

SpiralFlex · Aug 1, 2012

I assumed x was real.

Utility · Aug 1, 2012

Shouldn't it read 15 + 3i and x = 3 as the only solution?

SpiralFlex · Aug 1, 2012

Yup typo. Here we assume x is real. If it doesn't say then we need a different approach. Does it say x is real?

Utility · Aug 1, 2012

Good, that is the way I did it too. Thanks.

It doesn't say, but the solution was x = 3 in the book as well.