## tmybby Group Title what is a simpler form for this radical expression one year ago one year ago

1. tmybby

2. satellite73

the square root of 16 is 4 is a start

3. tmybby

4g^5?

4. satellite73

and half of 10 is 5, giving you $\sqrt{16g^{10}}=4g^5$

5. satellite73

yes you got it

6. tmybby

@satellite73 ?

7. satellite73

man they really reach for these word problems, don't they?

8. satellite73

perimeter is $\sqrt{13}+\sqrt{13}+7\sqrt{13}+7\sqrt{13}$ add to get $$16\sqrt{13}$$

9. satellite73

$$\sqrt{20}+\sqrt{40}-\sqrt{5}$$ you have to write the first two in simplest radical form do you know how to do that?

10. tmybby

no :/

11. satellite73

ok look to factor the number inside the radical as a perfect square, times something for example $$20=4\times 5$$ and $$4$$ is a square you get $\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\sqrt{5}=2\sqrt{5}$ you don't need to write all those steps, that is only my explanation

12. satellite73

see if you can try that with $$\sqrt{45}$$

13. tmybby

ohhh okay!

14. satellite73

if you get stuck let me know, but it is not too bad for $$45$$

15. tmybby

$\sqrt{45}=\sqrt{9x5}=\sqrt{9}\sqrt{5}$?

16. satellite73

yes !

17. satellite73

so finish with $$\sqrt{9}\sqrt{5}=3\sqrt{5}$$

18. satellite73

now the whole problem is a matter of combining like terms for $2\sqrt{5}+3\sqrt{5}-\sqrt{5}$ which is identical to $2+3-1$ really, just stick a $$\sqrt{5}$$ next to the answer

19. tmybby

$4\sqrt{5}$

20. satellite73

got it

21. satellite73

how about the last one $9^{\frac{1}{3}}\times 81^{\frac{1}{3}}$? you got that?

22. tmybby

I got 9 for the last one

23. satellite73

you win

24. tmybby

could you still help me?

25. satellite73

sure why not?

26. tmybby

27. satellite73

for a rational exponent, the power is the numerator and the root is the denominator

28. satellite73

your exponent will be negative, since the variable is in the denominator

29. tmybby

oh okay so it's C?

30. satellite73

should get $\large 8x^{-\frac{15}{7}}$ yes, C

31. satellite73

for the second one i can't really read the exponents

32. tmybby

33. satellite73

is the exponent in the numerator a 3 or an 8 ?

34. tmybby

3

35. satellite73

oh, then $$\sqrt[3]{x^3}=x$$ and you have $\frac{x}{\sqrt[5]{x^2}}$

36. satellite73

you can write this as $\frac{x}{x^{\frac{2}{5}}}$ and then subtract the exponents

37. satellite73

$\large x^{1-\frac{2}{5}}=x^{\frac{3}{5}}$

38. tmybby

yay! :)

39. satellite73

that it?

40. tmybby

i have 3 more problems /:

41. satellite73

i am ready if you are

42. tmybby

okay hold on... :)

43. tmybby

44. satellite73

just like the previous one, multiply top and bottom by $$\sqrt{2}+\sqrt{5}$$

45. satellite73

$\frac{\sqrt{2}+\sqrt{5}}{\sqrt2-\sqrt5}\times\frac{\sqrt{2}+\sqrt{5}}{\sqrt2+\sqrt5}$

46. satellite73

$\frac{2+2\sqrt{10}+5}{2-5}$ $\frac{7+2\sqrt{10}}{-3}$

47. tmybby

48. satellite73

any ideas for this one?

49. satellite73

one think should be ok, that $\large (a^{-3})^{\frac{-2}{3}}=a^{-3\times \frac{-2}{3}}=a^2$

50. satellite73

for $$8^{-\frac{2}{3}}$$ you have an exponent with a a) three in the deminator b) 2 in the numerator and c) a minus sign they mean a) take the cube root b) square c) take the reciprocal

51. satellite73

therefore you get the cubed root of 8 is 2 2 squared is 4 the reciprocal of 4 is $$\frac{1}{4}$$

52. satellite73

final answer is therefor $\frac{x^2}{4}$

53. tmybby

okay! One more...

54. tmybby

55. satellite73

that is an easier one multiply top and bottom by $$\sqrt3$$ to get $\frac{5\sqrt{3}+\sqrt{6}}{3}$

56. satellite73

typo there ! $\frac{5\sqrt{3}-\sqrt{6}}{3}$

57. tmybby

Thank you so much I really appreciate it!!

58. satellite73

yw