Question

How do you solve the quadratic with complex numbers given `3t^2+8t+5=-2t^2`?

Answer 1

`t=-4/5+3/5 i color(white)(..) and color(white)(..)t=-4/5-3/5 i`

Explanation:

Add `2t^2` to both sides

`5t^2+8t+5=0`

You can either complete the square or use the 'formula'

Keeping the standard form I will use `x`

`y=ax^2+bx+c =>x=(-b+-sqrt(b^2-4ac))/(2a)`

When I first came across these I made a point of writing the above at the start of every solution. It has stuck in my memory ever since. That was a long time ago so the method works!

`a=+5;b=+8 and c=+5` giving, in this case `t`:

`t=(-8+-sqrt(8^2-4(5)(5)))/(2(5))`

`t=-8/10+-sqrt(-36)/10`

Bit `-36` is the same as `36xx(-1)`

`t=-8/10+- [(sqrt(36)color(white)(.) sqrt(-1))/10]`

But `sqrt(36)=6 and sqrt(-1) = i`

`t=-4/5+-3/5 color(white)(.) i`

`t=-4/5+3/5 i color(white)(..) and color(white)(..)t=-4/5-3/5 i`

Answer 2

`t_1=-4/5+i3/5 or t_2=-4/5-i3/5`

Explanation:

`3t^2+8t+5=-2t^2|+2t^2`
`5t^2+8t+5=0|:5`
`t^2+8/5t+1=0`

Completing the square...

`(t+4/5)^2+1-16/25=0|-9/25|sqrt()`
`t+4/5=+-sqrt(-9/25)|-4/5`
`t=-4/5+-sqrt(-9/25)`

Determinating the complex root...

`sqrt(-a)=i*sqrt(a)`

`+-sqrt(-9/25)=+-i3/5`
`t_1=-4/5+i3/5 or t_2=-4/5-i3/5`