## anonymous 5 years ago Could anyone please help me with simplifying radicals? -15

1. anonymous

√(-15) is the problem?

2. anonymous

$\sqrt{-15}$

3. anonymous

what is your question?

4. anonymous

it is odd and cannot be simplfied

5. anonymous

ok so if you break it into it's factors you get √(-1)(3)(5) √(-1) = i the rest cannot be simplified so you have i√15 for an answer

6. anonymous

√49 is also odd but simplifies to 7 mmolina

7. anonymous

so the answer is -1/15

8. anonymous

?

9. anonymous

no, the answer is: i*srqr(15)

10. anonymous

i squared as in india type of i?

11. anonymous

$i*\sqrt{15}$

12. anonymous

hmm never heard of that before. what rule is that derived from?

13. anonymous

i is the imaginary unit √(-1) because you cannot take the square root of a negative number.

14. anonymous

okay. let me try that...

15. anonymous

set in you personal calculator $\sqrt{-1}$ this will give you an error as the answer. So every time you have an expresion like that you have to write an i in front of the root and make the number inside the root positive.

16. anonymous

the entire problem is quadratic formulas which is -1 +/- $\sqrt{-15}/2$

17. anonymous

then there are no real solutions

18. anonymous

I have the left side down and the fraction beneathe, but complicated fixing right side. is that possible for a solution set? the answer must be two separate numbers

19. anonymous

not always, quadratics have AT MOST 2 solutions but can have only one or no real solutions, this appears to be a classic no real solutions problem

20. anonymous

the solutions to a quadratic are where the graph crossed the x-axis, if you graph the original problem does it cross the x-axis?

21. anonymous

there is no graphing, and there is no option for no solution--there is indeed a solution. haha

22. anonymous

It might make things simpler if you tell us what the original problem is.

23. anonymous

it must form an equation with only addition and subraction signs between them

24. anonymous

I did if you look above

25. anonymous

:)

26. anonymous

hello?

27. anonymous

did you not understand it?

28. anonymous

I understand that you have simplified the original problem but if you post what the quadratic is then we could verify that your solutions are correct.