How do you evalute #\sqrt { 24} - 3\sqrt { 24} - 3\sqrt { 24}#?

2 Answers
Meave60
Feb 25, 2017

#sqrt(24) -3sqrt(24)-3sqrt(24)=-5sqrt(24)#

Explanation:

Each term contains #sqrt(24)#. This is similar to each term having the same variable. As long as they are the same, they can be added and subtracted directly.

#sqrt(24) -3sqrt(24)-3sqrt(24)#

The coefficient in front of the first term is understood to be #1#.

#1sqrt(24)-3sqrt(24)-3sqrt(24)=#

#-5sqrt(24)#

George C.
Feb 25, 2017

#sqrt(24)-3sqrt(24)-3sqrt(24) = -10sqrt(6) ~~ -24.4949#

Explanation:

First we can use the distributive property of multiplication over addition and subtraction to find:

#sqrt(24)-3sqrt(24)-3sqrt(24) = 1sqrt(24)-3sqrt(24)-3sqrt(24)#

#color(white)(sqrt(24)-3sqrt(24)-3sqrt(24)) = (1-3-3)sqrt(24)#

#color(white)(sqrt(24)-3sqrt(24)-3sqrt(24)) = -5sqrt(24)#

Next note that if #a, b >= 0# we have:

#sqrt(ab) = sqrt(a)sqrt(b)#

So we can simplify #sqrt(24)# as follows:

#sqrt(24) = sqrt(2^2*6) = sqrt(2^2)*sqrt(6) = 2sqrt(6)#

Putting these together, we have:

#sqrt(24)-3sqrt(24)-3sqrt(24) = -5sqrt(24) = -5*2sqrt(6) = -10sqrt(6)#

This cannot be simplified further, since #6# has no square factors.

#color(white)()#
Approximations

#sqrt(6)# is an irrational number somewhere between #sqrt(4) = 2# and #sqrt(9) = 3#.

If we want a rational approximation, then we can use a calculator to find:

#sqrt(6) ~~ 2.44948974#

or we can use any one of a number of methods to find it by hand.

#color(white)()#
For example, to find approximations to the square root of a number #n#, we can pick a first rational approximation #p_0/q_0# then repeatedly apply the following formulas:

#{ (p_(i+1) = p_i^2+n q_i^2), (q_(i+1) = 2p_i q_i) :}#

(This is an adaptation of the "Babylonian" method)

#color(white)()#
In our example, #n=6# and we can put #p_0/q_0 = 5/2# to find:

#{ (p_1 = p_0^2+n q_0^2 = 5^2+6*2^2 = 25+24 = color(blue)(49)), (q_1 = 2p_0 q_0 = 2*5*2 = color(blue)(20)) :}#

#{ (p_2 = p_1^2+n q_1^2 = 49^2+6*20^2 = 2401+2400 = color(blue)(4801)), (q_2 = 2p_1 q_1 = 2*49*20 = color(blue)(1960)) :}#

So:

#10sqrt(6) ~~ 10*4801/1960 = 4801/196 ~~ 24.4949#

So:

#sqrt(24)-3sqrt(24)-3sqrt(24) = -10sqrt(6) ~~ -24.4949#

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