## anonymous 5 years ago How do you find all the real solutions to x^3-13x-12=0?

1. anonymous

wait one moment

2. anonymous

$x \approx4 ; x \approx-3 ; x \approx-1$

3. anonymous

I was a bit complicated

4. anonymous

was rectified and that is the correct answer

5. anonymous

6. anonymous

wait one moment

7. anonymous

factor (x^2 -5)(x^2 -9)=0 x^2 = 5 x^2 = 9

8. anonymous

x=3 x=-3 x=square root(5) x=-square root(5)

9. anonymous

$\sqrt{5}\approx 2.23 ; -\sqrt{5}\approx - 2.23$

10. anonymous

Could you still factor it if the 14x^2 was 14x^3(in the same problem)?

11. anonymous

wait one moment

12. anonymous

is very complicated

13. anonymous

wait one moment

14. anonymous

Don't worry about it then. I was just wondering if they could be only one power different or if they could also be two powers away from each other.

15. anonymous

solve this equation is very long process and I think I'm doing that I have out there in 10%

16. anonymous

expected and it started to let me finish lol

17. anonymous

Can you show me the steps to the x^4-7x^3=0 problem? i think I know how to get 7 but I don't know how you got sero as an answer.

18. anonymous

only by deduction. in looking for possible solutions when the right side of the expression as a result of 0 in this case is very easy to see that 7 is the only number other than 0 that satisfies this condition and so that the other three answers are 0 and which is a 4-degree equation must give 4 answers

19. anonymous

Okay, so you get x^3(x-7)=0 and solve the two pieces for the answers?

20. anonymous

( x^3 ) (x-7)= 0 ??????

21. anonymous

I think that is how I was taught it. Then you get x=7 and x^3 =0