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

1. tmybby Group Title

2. satellite73 Group Title

the square root of 16 is 4 is a start

3. tmybby Group Title

4g^5?

4. satellite73 Group Title

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

5. satellite73 Group Title

yes you got it

6. tmybby Group Title

@satellite73 ?

7. satellite73 Group Title

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

8. satellite73 Group Title

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

9. satellite73 Group Title

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

10. tmybby Group Title

no :/

11. satellite73 Group Title

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 Group Title

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

13. tmybby Group Title

ohhh okay!

14. satellite73 Group Title

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

15. tmybby Group Title

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

16. satellite73 Group Title

yes !

17. satellite73 Group Title

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

18. satellite73 Group Title

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 Group Title

$4\sqrt{5}$

20. satellite73 Group Title

got it

21. satellite73 Group Title

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

22. tmybby Group Title

I got 9 for the last one

23. satellite73 Group Title

you win

24. tmybby Group Title

could you still help me?

25. satellite73 Group Title

sure why not?

26. tmybby Group Title

27. satellite73 Group Title

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

28. satellite73 Group Title

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

29. tmybby Group Title

oh okay so it's C?

30. satellite73 Group Title

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

31. satellite73 Group Title

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

32. tmybby Group Title

33. satellite73 Group Title

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

34. tmybby Group Title

3

35. satellite73 Group Title

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

36. satellite73 Group Title

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

37. satellite73 Group Title

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

38. tmybby Group Title

yay! :)

39. satellite73 Group Title

that it?

40. tmybby Group Title

i have 3 more problems /:

41. satellite73 Group Title

i am ready if you are

42. tmybby Group Title

okay hold on... :)

43. tmybby Group Title

44. satellite73 Group Title

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

45. satellite73 Group Title

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

46. satellite73 Group Title

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

47. tmybby Group Title

48. satellite73 Group Title

any ideas for this one?

49. satellite73 Group Title

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

50. satellite73 Group Title

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 Group Title

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

52. satellite73 Group Title

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

53. tmybby Group Title

okay! One more...

54. tmybby Group Title

55. satellite73 Group Title

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

56. satellite73 Group Title

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

57. tmybby Group Title

Thank you so much I really appreciate it!!

58. satellite73 Group Title

yw