Question 1094775
THE WORD PROBLEM PART:
{{{sqrt(3)-sqrt(2)}}}= length of a side of the square base in meters.
{{{(sqrt(3)-sqrt(2))^2=3+2-2sqrt(3)sqrt(2)=5-2sqrt(3)sqrt(2)=5-2sqrt(6)}}}= surface area of the square base in square meters.
With that and the {{{4sqrt(2)-3sqrt(3)}}} volume of the block in cubic meters,
knowing that volume = (area of the base)(height),
we can calculate the height in  meters as
{{{(4sqrt(2)-3sqrt(3))/(5-2sqrt(3)sqrt(2))}}} or {{{(4sqrt(2)-3sqrt(3))/(5-2sqrt(6))}}} .
 
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 {{{(5+2sqrt(6))}}} ,
because {{{(5-2sqrt(6))(5+2sqrt(6))=5^2-(2sqrt(6))^2-25-2^2(sqrt(6))^2=25-4*6=25-24-1}}} .
{{{(4sqrt(2)-3sqrt(3))(5+2sqrt(6))/(5-2sqrt(6))/(5+2sqrt(6))}}} or {{{(4sqrt(2)-3sqrt(3))(5+2sqrt(3)sqrt(2))/(5-2sqrt(3)sqrt(2))/(5+2sqrt(3)sqrt(2))}}}
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
{{{(4sqrt(2)-3sqrt(3))(5+2sqrt(6))}}}{{{"="}}}{{{20sqrt(2)+4sqrt(2)*2sqrt(6)-15sqrt(3)-3sqrt(3)*2sqrt(6)}}}{{{"="}}}{{{20sqrt(2)+8sqrt(12)-15sqrt(3)-6sqrt(18)}}}{{{"="}}}{{{20sqrt(2)+8sqrt(4*3)-15sqrt(3)-6sqrt(9*2)}}}{{{"="}}}{{{20sqrt(2)+8*2sqrt(3)-15sqrt(3)-6*3sqrt(2)}}}{{{"="}}}{{{20sqrt(2)+16sqrt(3)-15sqrt(3)-18sqrt(2)}}}{{{"="}}}{{{highlight(2sqrt(2)+sqrt(3))}}} .
Or you could calculate it as
{{{(4sqrt(2)-3sqrt(3))(5+2sqrt(3)sqrt(2))}}}{{{"="}}}
{{{20sqrt(2)+4sqrt(2)*2sqrt(3)sqrt(2)-15sqrt(3)-3sqrt(3)*2sqrt(3)sqrt(2)}}}{{{"="}}}
{{{20sqrt(2)+4*2*2sqrt(3)-15sqrt(3)-3*2*3sqrt(2)}}}{{{"="}}}
{{{20sqrt(2)+16sqrt(3)-15sqrt(3)-18sqrt(2)}}}{{{"="}}}
{{{highlight(2sqrt(2)+sqrt(3))}}} .
 
VERIFYING YOUR ANSWER:
You could also calculate approximate values for
{{{4sqrt(2)-3sqrt(3)}}}{{{"="}}}{{{approximately}}}{{{0.460702}}} and
{{{5+2sqrt(6)}}}{{{"="}}}{{{approximately}}}{{{9.898979}}} ,
and then multiply them together to find the height in meters as
{{{approximately}}}{{{0.460702*9.898979}}}{{{"="}}}{{{approximately}}}{{{4.560478}}} .

That is not the answer "in the form of {{{a*sqrt(2)+b*sqrt(3)}}} ,"
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:
{{{(2sqrt(2)+sqrt(3))*(5-2sqrt(6))=10sqrt(2)-4sqrt(12)+5sqrt(3)-2sqrt(18)}}}{{{"="}}}
{{{10sqrt(2)-4sqrt(4*3)+5sqrt(3)-2sqrt(9*2)=10sqrt(2)-8sqrt(3)+5sqrt(3)-6sqrt(2)}}}
{{{"="}}}{{{4sqrt(2)-3sqrt(3)}}} .
 
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+sqrt(5)}}} and {{{7-sqrt(5)}}} , and
3. the idea of simplifying roots as in {{{2sqrt(18)=2sqrt(9*2)=2sqrt(9)*sqrt(2)=2*3*sqrt(2)=6sqrt(2)}}} .