Question

Need help with quadratic equations
Solve: -3x^3 - 4x - 4 = 0

Answer 1

which methods are you familiar with? can you factor? use the quadratic equation? complete the square?

Answer 2

I'm familiar with the quadratic equation

Answer 3

so why don't you use that?

Answer 4

ooh, sorry... didn't see the cube.

is it \[-3x^3-4x-4=0\]

Answer 5

or is it \[-3x^2-4x-4=0\]

Answer 6

No its squared and when i do it i get up to 4+/-sqrt(-32)/-6 and get stuck

Answer 7

the second one

Answer 8

so here's the deal... "can't" take the square root of a negative number, right?

well, turns out you can.

we call \(i\) the imaginary number. \(i=\sqrt{-1}\).

so anytime you have to take the square root of a negative number, you factor out \(i\).

for example: suppose you have to take the square root of -16...\[\sqrt{-16}=\sqrt{16}\cdot \sqrt{-1} = 4\cdot i = 4i\]

Answer 9

it is customary to put rational numbers (integers and fractions) before \(i\) and irrational numbers (square, cube, etc. roots) after \(i\). That's why I wrote it as \(4i\) and not \(i4\).

Answer 10

oh ok, could you help me simplify it more? because its a practice test and the answers are

Answer 11

http://imgur.com/RlyOSLp

Answer 12

\[\text{When }ax^+bx+c=0 \text{, then }\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a } \Rightarrow \frac{ -\left( -4 \right) \pm \sqrt{\left( -4 \right)^2-4\left( -3 \right) \left( -4 \right)} }{ 2\left( -3 \right) }\]\[\Rightarrow \frac{ 4 \pm \sqrt{16-48} }{ -6 }=\frac{ 4 \pm \sqrt{-32} }{ -6 }=\frac{ 4 \pm \sqrt{16\cdot2}\cdot \sqrt{-1} }{ -6 }=\frac{ 4 \pm 4i \sqrt{2} }{ -6 }=-\frac{ 2 }{3 }\pm\frac{ 2i }{3 }\]

Answer 13

or as they have it...\[=\frac{ -2\pm 2i \sqrt{2} }{3 }\]

Answer 14

learn to do this work yourself and you WILL get better at it. use this site to have people check your work and help with explanations.

Answer 15

Oh i see what you do now, thank you so much for your help!