Question

x^3 + 4x^2 + 21 = 0 Solving for x

Answer 1

Can you do the start?

Answer 2

I have know idea where to begin. Do you subtract 21 from both sides?

Answer 3

Are you really solving \[x^3+4x^2+21=0\] and not \[x^2+4x+21=0\]?

Answer 4

I'm sure.

Answer 5

What class is this for?

Answer 6

Algebra 2

Answer 7

but one of the exponents is 3, not 2

Answer 8

I can solve a regular quadratic, but I don't know how to do this one.

Answer 9

Oh, I see. :/ @Mertsj @Luis_Rivera

Answer 10

I can tell you what the answers are, but not how to get them:

Here's the real root:
\[\frac{1}{3} \left(-4-16 \left(\frac{2}{695-9 \sqrt{5761}}\right)^{1/3}-\left(\frac{1}{2} \left(695-9 \sqrt{5761}\right)\right)^{1/3}\right)\]

That's why I'm skeptical that you're really supposed to be solving a cubic rather than a quadratic...

Answer 11

Btw, @jazy, that should be \[ax^2+bx+c=0\] (same letters you use in the formula, and the middle term corrected)

Answer 12

Right^

Answer 13

thanks anyway