## blast234 Group Title .... one year ago one year ago

1. Laura<3 Group Title

Uh... Sorry. I don't know. What math are you taking?

2. phi Group Title

the first step might be to use this rule $\frac{\sqrt{a}}{\sqrt{b} } = \sqrt{\frac{a}{b}}$

3. phi Group Title

that is a good way to get rid of the sqrt in the bottom. you can then use $\sqrt{a}\cdot \sqrt{b}= \sqrt{a \cdot b}$ to merge the 2 sorts then find the prime factors of the number, and "pull out" pairs of the same number from the sort for the x's pull out pairs of x's (remember x^16 is x*x*x...*x (x times itself 16 times, so lots of pairs)

4. ximki Group Title

125x^15

5. phi Group Title

Let's do it this way $\frac{\sqrt{250 x^{16}}}{\sqrt{2x}}= \sqrt{\frac{250 x^{16}}{2x}}$ simplify to get rid of the fraction. divide 2 and x into the top. what do you get?

6. phi Group Title

write out the whole answer, with the sort

7. phi Group Title

what happened to the x in the bottom ? $\sqrt{\frac{250 x^{16}}{2x}}$

8. phi Group Title

remember x*x*x/x is x*x one x from the top cancels out the x in the bottom

9. phi Group Title

we start with $\sqrt{\frac{250 x^{16}}{2x}}$ now divide 2 into 250. you get 125 divide x into x^16 . what do you get?

10. phi Group Title

x^16 is short hand for x*x*x*x*x*x*x*x*x*x*x*x*x*x*x*x if you divide that by x what do you get?

11. phi Group Title

yes. Let's do a simple example $\frac{x}{x}$ that is 1 $\frac{x\cdot x}{x} = \frac{\cancel{x}\cdot x}{\cancel{x}} =x$

12. phi Group Title

In your problem we cancel out 1 x from the top and bottom that leaves 15 x's multiplied together. we would use short hand to write it x^15

13. phi Group Title

closer but when you divide, you don't have the denominator anymore. in other words, we had 250/2 we get 125 NOT 125/2 and x*x/x is x NOT x/x can you fix your answer?

14. phi Group Title

yes now let's do the easy part: the x^15 Here is an example with x^2: $\sqrt{x\cdot x} = x$ we "pull out" a pair of x's and make it just x with no square root Here is an example with x^3 $\sqrt{x\cdot x\cdot x} = x\sqrt{x}$ pull out a pair of x's , but leave the last one inside the square root what do we do with x^15 ?

15. phi Group Title

yes, but we can think about it like this. 15 is an odd number, but 14 is even 14 is 2*7 so we see there are 7 pairs of x's and 1 left over to make 15 $\sqrt{x^{15}} = x^7 \sqrt{x}$

16. phi Group Title

now the 125. we factor it into prime numbers, and look for pairs. 5 divides into 125. can you find the factors?

17. phi Group Title

125 is 5*5*5 NO 7 involved. Can you simplify $\sqrt{5 \cdot 5 \cdot 5}$?

18. phi Group Title

you can "pull out" one pair of 5's

19. phi Group Title

except when the pair comes out, you drop one of the 5's

20. blast234 Group Title

is that right

21. phi Group Title

yes, now put it all together.

22. phi Group Title

$\sqrt{125 x^{15} }$ is ?

23. blast234 Group Title

How can i put them together

24. phi Group Title

$\sqrt{125 x^{15} } = \sqrt{125} \cdot \sqrt{x^{16}}$ and you know the answer to each of the square roots (we just figured them out)

25. phi Group Title

that is the simplified sort(125), but you need to multiply it by sqrt(x^15) (simplified)

26. phi Group Title

sqrt(125) * sqrt(x^15) = 5 * sqrt(5) * x^7 * sqrt(x) we can combine the sqrt(5)*sqrt(x) into sqrt(5x) so the final answer is 5 x^7 sqrt(5x)

27. phi Group Title

Try some more, because it takes practice.

28. blast234 Group Title

Ok, i will