Calculations?

Question

Hi.
I have some problems with sqrt and I am hoping someone can help me. Could someone please walk me through how to solve these step by step.
1) 2 times sqrt a^5 minus a times sqrt a^3
2) sqrt 98 minus sqrt 128 divided by sqrt 8 minus sqrt 2
3) sqrt 3 divided by sqrt 27 plus sqrt 3
Thank you.

Answer 1

1) 2*sqrt(a^5) - a*sqrt(a^3)
= 2*sqrt(a^4)*sqrt(a) - a*sqrt(a^2)*sqrt(a)
= 2a^2*sqrt(a) - a^2*sqrt(a)
= a^2*sqrt(a)

2) sqrt(98) - sqrt(128) / sqrt(8) - sqrt(2)
= 7*sqrt(2) - 8*sqrt(2) / 2*sqrt(2) - sqrt(2)
= 7*sqrt(2) - 4 - sqrt(2)
= 6*sqrt(2) - 4

3) sqrt(3) / sqrt(27) + sqrt(3)
= sqrt(3) / 3*sqrt(3) + sqrt(3)
= 1/3 + sqrt(3)

Answer 2

2 * sqrt(a^5) - a * sqrt(a^3) = ....

i dont know what you want to do with these

somewthing you can do is :

replace the sqrt by ^0.5

2 * sqrt(a^5) - a * sqrt(a^3) = ....
2 * a^(5/2) - a * a^(3/2) =
2 * a^(5/2) - a^(2/2) * a^(3/2) =
2 * a^(5/2) - a^(5/2) =
a^(5/2)
ahhh ok, make the powers of the numbers equal if possible then add / subtract he things,
-----------------------
[ sqrt (98) - sqrt (128) ] / [ sqrt (8) - sqrt (2) ]

98 = 2 * 49 = 2 * 7 * 7
128 = 2 ^ 7
so sqrt(98) = 7 * sqrt(2) = 7 * 2 ^(1/2)
sqrt(128) = sqrt(2 ^ 7) = 2 ^(7/2) = 2^3*2^(1/2) = 8 * 2^(1/2)

so sqrt(98) - sqrt(128) = 7 * 2 ^(1/2) - 8 * 2^(1/2) = - 2^(1/2) = - sqrt(2)

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hope this helps enough ?

Answer 3

1) i'm not sure exactly what this equation is:
2(sqrt a^5) - (a)(sqrt a^3) perhaps?
anyway, looking at sqrt a^3...
sqrt of a number can also be written as a^(1/2)
[ a to the one half power]
so sqrt a^3 = a ^(3/2)

2) you need to show the equations in mathematical notation rather than writing it out.
sqrt 98 = sqrt(49)(2) = 7sqrt2
sqrt 128 = sqrt(64)(2) = 8sqrt2
sqrt 8 = sqrt(4)(2) = 2sqrt2
can you take it from here?