# Thread: How would you solve 7x^2-10x=4x^3-9?

1. ## How would you solve 7x^2-10x=4x^3-9?

7x2-10x=4x3-9
x=?

2. ## Re: How would you solve 7x^2-10x=4x^3-9?

Originally Posted by EJdive43
7x2-10x=4x3-9
x=?
$\displaystyle 4x^3-7x^2+10x-9=0$
There is no 'nice' answer.

3. ## Re: How would you solve 7x^2-10x=4x^3-9?

But I got x= 7/4,9, and 1

4. ## Re: How would you solve 7x^2-10x=4x^3-9?

Originally Posted by EJdive43
But I got x= 7/4,9, and 1
None of that is correct.

5. ## Re: How would you solve 7x^2-10x=4x^3-9?

1)I thought you bring everything to one side: -4x^3+7x^2-10x-9=0
2) then you factor: x^2(-4x+7)-10x+9=0
3) then set everything to zero: (-4x+7)=0 x^2-10x+9=0
4) then solve: X=7/4. X=9 x=1
Is that not right? What did I do wrong? >_<

6. ## Re: How would you solve 7x^2-10x=4x^3-9?

Originally Posted by EJdive43
1)I thought you bring everything to one side: -4x^3+7x^2-10x-9=0
2) then you factor: x^2(-4x+7)-10x+9=0
3) then set everything to zero: (-4x+7)=0 x^2-10x+9=0
4) then solve: X=7/4. X=9 x=1
Is that not right? What did I do wrong? >_<
Here is the solution.

7. ## Re: How would you solve 7x^2-10x=4x^3-9?

Originally Posted by EJdive43
1)I thought you bring everything to one side: -4x^3+7x^2-10x-9=0
2) then you factor: x^2(-4x+7)-10x+9=0
3) then set everything to zero: (-4x+7)=0 x^2-10x+9=0 this step is incorrect ... one sets factors equal to zero, not terms.
4) then solve: X=7/4. X=9 x=1
Is that not right? What did I do wrong? >_<
...

8. ## Re: How would you solve 7x^2-10x=4x^3-9?

If your really believe that 7- 10= 4- 9, then your first mistake was not paying attention in arithmetic!