# Solve for x: 2x-x^2=3

1. Aug 28, 2011

### davie08

1. The problem statement, all variables and given/known data

solve for x: 2x-x^2=3

2. Relevant equations

3. The attempt at a solution

2x-x^2=3

do i go -x^2=3-2x

thats not right i can tell but i got a warning for not showing work and for this question again i dont know where to start but maybe if i throw a guess in this will satisfy the guy who watches for this.

actually this is too much of a hassle since im doing so many questions in such a short time period i will be back in another week when i start my actual math course sorry mark im just stressed over doing 8 hours of math everyday.

2. Aug 28, 2011

### slain4ever

make the equation = 0
then either use the quadratic formula or factorising to solve for x

3. Aug 28, 2011

### Staff: Mentor

Move all of the terms over to one side, so that zero is on the other side. Starting from your first equation,

2x-x^2=3

rewrite this as
-x^2 + 2x - 3 = 0

You can make the x^2 coefficient positive by multiplying both sides by -1.

4. Aug 28, 2011

### davie08

okay I think I will just learn the quad formula for this so with making the x^2 positive would I make the 2x negative and the -3 positive.

5. Aug 28, 2011

### davie08

I end up with a -8 in a square root when I use the quad formula what would I do from there or did I screw up somewhere.

6. Aug 28, 2011

### slain4ever

yep there should be a negative number in the square root, which means there are no x solutions

ie. the graph of the function does not cross the x axis.

so no real solutions of course there are 'imaginery' solutions

7. Aug 28, 2011

### davie08

could I use the quad. formula for a question like 0=125+27k^3

8. Aug 28, 2011

### slain4ever

no the quardratic formula is for something to the power of 2 not 3 or more
there is a forumula for the power of 3 but it's quite complicated. It is easier to factor out something to end up with a power of 2.
in this case you can factor out 3x+5

9. Aug 28, 2011

### davie08

for 0= 125+27k^3 would the answer be k=5/3

10. Aug 28, 2011

### ArcanaNoir

The formula for factoring the sum of cubes (a^3+b^3) is (a+b)(a^2-2ab+b^2)
so, (5+3k)(25-30k+9k^2)=0
so one solution is found by 5+3k=0
and another by 25-30k+9k^2=0

11. Aug 29, 2011

### HallsofIvy

Staff Emeritus
The simplest way to solve $125- 27k^3= 0$ is to write it as $27k^3= 125$, divide both sides by 27, and take the cube root of each side. As AracnNoir points out, there are two other solutions (the roots of his $25- 30k+ 9k^2= 0[/tex] but they are both complex numbers. Yes, [itex]125= 5^3$ and $27= 3^3$ so 5/3 is an answer. You could have checked that yourself- if x= 5/3, then $x^3= 5^3/3^3= 125/27$ so 27x^3= 27(125/27)= 125.