Adding and subtracting Radicals?

Question

-4 √(75) +3√(147)

3√(5/12) + √(12/5) -1/3√60

Answer 1

First, factor the numbers in the radicals to extract perfect square factors:

-4√(25*3) + 3√(49*3)

Break the roots up:

-4√25√3 + 3√49√3

Simplify:

-4*5√3 + 3*7√3
-20√3 + 21√3

Now combine like terms:

(-20+21)√3
√3

For the second one, first break up the quotients in the square roots:

3√5/√12 + √12/√5 - √60/3

Find a common denominator:

3√25/√60 + √144/√60 - √3600/(3√60)

I'm not quite finding a common denominator yet, since I know the 3 will cancel. Simplify the square roots:

15/√60 + 12/√60 - 60/(3√60)

Now, cancel the factor of 3:

15/√60 + 12/√60 - 20/√60

Combine like terms:

(15+12-20)/√60
7/√60

We can simplify this further. Extract all square factors from the denominator:

7/(2√15)

And if your teacher is a stickler about not having square roots in the denominator:

7√15/(2*15)
7√15/30

And we are done.

Answer 2

sqrt 75=5*sqrt3
sqrt 147=7*sqrt3
-20sqrt3+21sqrt3=sqrt3

3sqrt(5*5/60) + sqrt (12*12/60)-1/3 sqrt60 **you multiply the first fraction with sqrt5 and the second with sqrt 12 so that you obtain sqrt60
15/sqrt60+12/sqrt60-sqrt60/3 **now you'll have to multiply the third fraction(-sqrt60/3) with sqrt60 so that you obtain sqrt60*sqrt60(=60) divided by 3*sqrt60,which simplified becomes 20/sqrt60
27/sqrt 60-20/sqrt60=7/sqrt60

Answer 3

-4 √(75) +3√(147)
= -4 √(25)√(3) +3√(49)√(3)
= -4*5√(3) + 3*7√(3)
= -20√(3) + 21√(3)
= √(3)

Answer 4

What you ought to do is shrink the sqrt's to being a similar. for occasion, Sqrt(18) = sqrt(2 * 9) = sqrt(9) * sqrt(2) = 3 sqrt(2) so, your first expression will become 4 * sqrt(2) + 18 * sqrt(2) = 22 * sqrt(2) the 2nd works similarly (40 8 = sixteen * 3 and seventy 5 = 25 * 3 ==> sixteen = 4^2 and 25 =5^2) so 5 * 4 * sqrt(3) - 4 * 5 sqrt(3) = 0

Answer 5

-4 sq rt (25x3) + 3sq rt(49x3)=
-4(5) sq rt 3 + 3(7) sq rt 3=
-20 sq rt 3 + 21 sqrt 3 = 1 sq rt 3


Mult by sq rt (60/60)

Answer 6

the radicals have to have similar root values before you subtract or add them