# Math Help - Quadratic equation check

I can't seem to figure out how to check this problem.

8x^3 + 1 =0

I factored it and got

(2x+1)(4x^2 - 2x + 1)

So x=-1/2 and x= (1±i sqrt 3)/4

The first solution -1/2 seems to satisfy the equation, but I can't figure out how to check for the second solution.

8[(1±i sqrt 3)/4]^3 +1=0

I multiplied the first two numbers and got
(-2±i sqrt 3)/2 + 1

and then I got stuck...

So, could someone please tell me where to go from here?

Any help would be greatly appreciated.

Thanks

2. For the second solution

Actually there are 3 solutions:
The "second" solution composed of two:
[1 +i*sqrt(3)]/4
and [1 -i*sqrt(3)]/4

It's complicated to check with the plus-or-minus, so let's check with the plus,
x = [1 +i*sqrt(3)]/4

8x^3 +1 = 0
8{[1 +sqrt(-3)]/4}^3 +1 =? 0

We can use the formula (a+b)^3 = a^3 +3(a^2)(b) +3(a)(b^2) +b^3
but that is complicated too, or very long.
So we use (a+b)^3 = (a+b)(a+b)^2

{[1 +sqtrt(-3)]/4}^2
= (1/16){1 +2sqrt(-3) -3}
= (1/16){-2 +2sqrt(-3)}
= (1/8){-1 +sqrt(-3)}

Multiply that by [1 +sqrt(-3)]/4,
= (1/8){-1 +sqrt(-3)} * [1 +sqrt(-3)]/4
= (1/32){-1*1 -1*sqrt(-3) +sqrt(-3) -3}
= (1/32){-4}
= -1/8

So,
8x^3 +1 = 0
8{[1 +sqrt(-3)]/4}^3 +1 =? 0
8{-1/8} +1 =? 0
-1 +1 =? 0
0 =? 0
Yes, so, OK, [1 +i*sqrt(3)]/4 is a root.

Try the x = [1 -i*sqrt(3)]/4.
It should check too.

3. Originally Posted by mi986
I can't seem to figure out how to check this problem.

8x^3 + 1 =0

I factored it and got

(2x+1)(4x^2 - 2x + 1)

So x=-1/2 and x= (1±i sqrt 3)/4

The first solution -1/2 seems to satisfy the equation, but I can't figure out how to check for the second solution.

8[(1±i sqrt 3)/4]^3 +1=0

I multiplied the first two numbers and got
(-2±i sqrt 3)/2 + 1

and then I got stuck...

So, could someone please tell me where to go from here?

Any help would be greatly appreciated.

Thanks
$
(1+ i \sqrt{3})^3 = 1 + 3(\pm i \sqrt{3})+ 3(\pm i \sqrt{3})^2+(\pm i \sqrt{3})^3
$

.............. $
= 1 \pm 3 i \sqrt{3} - 9 \mp 3 i \sqrt{3} = -8
$

RonL

4. Ah, thank you both so much. I figured out where I went wrong! Instead of $
\mp
$

I used ± both times when I cubed $
1 \pm i \sqrt{3}$