SOLUTION: A rectangular block has a square base. The length of each side of the base is (√3 - √2)m and volume of the block is (4√2 - 3√3) cubic metre. Find the height

Algebra ->  Square-cubic-other-roots -> SOLUTION: A rectangular block has a square base. The length of each side of the base is (√3 - √2)m and volume of the block is (4√2 - 3√3) cubic metre. Find the height      Log On


   



Question 1094775: A rectangular block has a square base. The length of each side of the base is (√3 - √2)m and volume of the block is (4√2 - 3√3) cubic metre. Find the height of the block in the form (a√2 + b√3) where a and b are constants.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE WORD PROBLEM PART:
sqrt%283%29-sqrt%282%29= length of a side of the square base in meters.
= surface area of the square base in square meters.
With that and the 4sqrt%282%29-3sqrt%283%29 volume of the block in cubic meters,
knowing that volume = (area of the base)(height),
we can calculate the height in meters as
%284sqrt%282%29-3sqrt%283%29%29%2F%285-2sqrt%283%29sqrt%282%29%29 or %284sqrt%282%29-3sqrt%283%29%29%2F%285-2sqrt%286%29%29 .

HOW TO SIMPLIFY THAT QUOTIENT:
To get rid of the irrationality in the denominator,
you need to do something you probably do almost often: rationalize.
In math, you do that by multiplying numerator and denominator times the irrational number that will result in a rational denominator.
In thid case, the irrational factor we need to multiply by is %285%2B2sqrt%286%29%29 ,
because .
or
With either of those equivalent expressions, the denominator is 1 ,
and all you have to do is carefully calculate the numerator,
without making mistakes.
You could calculate it as
%284sqrt%282%29-3sqrt%283%29%29%285%2B2sqrt%286%29%29%22=%2220sqrt%282%29%2B4sqrt%282%29%2A2sqrt%286%29-15sqrt%283%29-3sqrt%283%29%2A2sqrt%286%29%22=%2220sqrt%282%29%2B8sqrt%2812%29-15sqrt%283%29-6sqrt%2818%29%22=%2220sqrt%282%29%2B8sqrt%284%2A3%29-15sqrt%283%29-6sqrt%289%2A2%29%22=%2220sqrt%282%29%2B8%2A2sqrt%283%29-15sqrt%283%29-6%2A3sqrt%282%29%22=%2220sqrt%282%29%2B16sqrt%283%29-15sqrt%283%29-18sqrt%282%29%22=%22highlight%282sqrt%282%29%2Bsqrt%283%29%29 .
Or you could calculate it as
%284sqrt%282%29-3sqrt%283%29%29%285%2B2sqrt%283%29sqrt%282%29%29%22=%22
%22=%22
20sqrt%282%29%2B4%2A2%2A2sqrt%283%29-15sqrt%283%29-3%2A2%2A3sqrt%282%29%22=%22
20sqrt%282%29%2B16sqrt%283%29-15sqrt%283%29-18sqrt%282%29%22=%22
highlight%282sqrt%282%29%2Bsqrt%283%29%29 .

VERIFYING YOUR ANSWER:
You could also calculate approximate values for
4sqrt%282%29-3sqrt%283%29%22=%22approximately0.460702 and
5%2B2sqrt%286%29%22=%22approximately9.898979 ,
and then multiply them together to find the height in meters as
approximately0.460702%2A9.898979%22=%22approximately4.560478 .
That is not the answer "in the form of a%2Asqrt%282%29%2Bb%2Asqrt%283%29 ,"
but it is easy to calculate with a computer or calculator,
and a way to verify if you made a mistake.

Another way to verify the answer
is to multiply area in square meters = times the answer you found:
%22=%22

%22=%224sqrt%282%29-3sqrt%283%29 .

NOTE: I have not yet decided if assigning this problem is a way to give students practice with square roots, or a way to torture them. I suppose that assigning one or two such exercises would drive home
1. the need to understand the situation in word problems to figure out the needed calculations ,
2. the idea of "rationalizing" denominators by multiplying "conjugate irrational numbers" as a pair of irrational numbers like 7%2Bsqrt%285%29 and 7-sqrt%285%29 , and
3. the idea of simplifying roots as in 2sqrt%2818%29=2sqrt%289%2A2%29=2sqrt%289%29%2Asqrt%282%29=2%2A3%2Asqrt%282%29=6sqrt%282%29 .