If only it were something easy like sqrt(25) or something simple that has a nice round square root since that's like saying sqrt(5*5). Well looking at the form... maybe it can be factored into something pretty inside the square root... something like sqrt( a*a) so that when we square root it, we can remove one of those a's and have the single a by itself. Well it looks like we are adding two things inside the square root so maybe instead of a*a, we have something along the lines of (a+b)*(a+b) which is the same as (a+b)^2. Yeah, if we can get the part inside the first big radical into the form (a+b)^2 then the square root strips off the squared and we're left with the simple (a+b) part coming out.
Well let's call that sqrt((a+b)^2) where (a+b)^2 = 3+2*sqrt(2). Now we just need to figure out what a and b are. Once we know that, we can know that the original expression simplifies to just (a+b) instead of sqrt((a+b)^2).
What is our motivation for this? Well if it turns out that the radicand (what's under that first radical) is some simple product of a+b squared... then when we square root it, we'll be left with the simple a+b and we'll have removed one of the radicals.
Well let's expand out this (a+b)^2 and set it equal to our radicand and see if we can't determine what a and b should be to make this little trick work...
a^2+2*a*b+b^2 = 3+2*sqrt(2)
Now let's let a = sqrt(3)... This is really just our choice. The first term is a^2 and we have a nice 3 sitting there so let's pretend that a^2 = 3 (such that a = sqrt(3)) We can then figure out what b would need to be to make our (a+b)^2 expression still equal the original radicand (3+2*sqrt(2))
Substituting in a=sqrt(3) into the equation of a^2+2*a*b+b^2 = 3+2*sqrt(2) we get the following...
sqrt(3)^2 + 2*sqrt(3)*b+b^2 = 3+2*sqrt(2)
Let's combine terms...
3+2*sqrt(3)*b+b^2 = 3+2*sqrt(2)
Let's now rearrange it into the quadratic form c1*b^2 + c2*b + c3 = 0. notice I had the variable b here and so I didn't want to confuse you by calling the coefficicnets by a,b and c although you may more easily recognize a quadratic equation as a*x^2+b*x+c = 0.
b^2 + 2*sqrt(3)*b - 2*sqrt(2) = 0
Quadratic equation tells us that b would therefore need to be... (solve using quadratic equation)
b = -sqrt(3) + sqrt(2) +1
or
b = -sqrt(3) - sqrt(2) -1
So a=3, b = the answers above. And if those add then square to give the original radicand (a+b)^2, we can keep (a+b) as the simplified version after taking the square root of (a+b)^2 to strip off that squared part.
Our expression of (a+b), the simplified result, is found to be:
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Basically you need to know sqrt(3+2*sqrt(2)).
If only it were something easy like sqrt(25) or something simple that has a nice round square root since that's like saying sqrt(5*5). Well looking at the form... maybe it can be factored into something pretty inside the square root... something like sqrt( a*a) so that when we square root it, we can remove one of those a's and have the single a by itself. Well it looks like we are adding two things inside the square root so maybe instead of a*a, we have something along the lines of (a+b)*(a+b) which is the same as (a+b)^2. Yeah, if we can get the part inside the first big radical into the form (a+b)^2 then the square root strips off the squared and we're left with the simple (a+b) part coming out.
Well let's call that sqrt((a+b)^2) where (a+b)^2 = 3+2*sqrt(2). Now we just need to figure out what a and b are. Once we know that, we can know that the original expression simplifies to just (a+b) instead of sqrt((a+b)^2).
What is our motivation for this? Well if it turns out that the radicand (what's under that first radical) is some simple product of a+b squared... then when we square root it, we'll be left with the simple a+b and we'll have removed one of the radicals.
Well let's expand out this (a+b)^2 and set it equal to our radicand and see if we can't determine what a and b should be to make this little trick work...
a^2+2*a*b+b^2 = 3+2*sqrt(2)
Now let's let a = sqrt(3)... This is really just our choice. The first term is a^2 and we have a nice 3 sitting there so let's pretend that a^2 = 3 (such that a = sqrt(3)) We can then figure out what b would need to be to make our (a+b)^2 expression still equal the original radicand (3+2*sqrt(2))
Substituting in a=sqrt(3) into the equation of a^2+2*a*b+b^2 = 3+2*sqrt(2) we get the following...
sqrt(3)^2 + 2*sqrt(3)*b+b^2 = 3+2*sqrt(2)
Let's combine terms...
3+2*sqrt(3)*b+b^2 = 3+2*sqrt(2)
Let's now rearrange it into the quadratic form c1*b^2 + c2*b + c3 = 0. notice I had the variable b here and so I didn't want to confuse you by calling the coefficicnets by a,b and c although you may more easily recognize a quadratic equation as a*x^2+b*x+c = 0.
b^2 + 2*sqrt(3)*b - 2*sqrt(2) = 0
Quadratic equation tells us that b would therefore need to be... (solve using quadratic equation)
b = -sqrt(3) + sqrt(2) +1
or
b = -sqrt(3) - sqrt(2) -1
So a=3, b = the answers above. And if those add then square to give the original radicand (a+b)^2, we can keep (a+b) as the simplified version after taking the square root of (a+b)^2 to strip off that squared part.
Our expression of (a+b), the simplified result, is found to be:
a+b = [sqrt(3)] + [-sqrt(3) + sqrt(2) +1] = sqrt(2)+1
or
a+b = [sqrt(3)] + [- sqrt(3) -sqrt(2)-1] = -(sqrt(2)+1)
(Remember we got two solutions from the quadratic solution for b, so we have two solutions here.
So our answer is ±(sqrt(2)+1)
And of course that makes sense that we have a positive or negative version since you get that when you take a square root.
Suppose that:
sqrt(x) + sqrt(y) = sqrt(3 + 2*sqrt(2))
Square both sides:
x + 2*sqrt(x)*sqrt(y) + y = 3 + 2*sqrt(2)
(x + y) + 2*sqrt(xy) = 3 + 2*sqrt(2)
Therefore we have two equations:
x + y = 3
xy = 2
From the first equation, y = 3 - x
Substitute that into the second equation:
x(3 - x) = 2
3x - x^2 = 2
-x^2 + 3x - 2 = 0
x^2 - 3x + 2 = 0
x^2 - 2x - x + 2 = 0
x(x - 2) - (x - 2) = 0
(x - 1)(x - 2) = 0
x - 1 = 0 or x - 2 = 0
x = 1 or x = 2
If x = 1 then y = 2/1 = 2
If x = 2 then y = 2/2 = 1
So the answer is:
sqrt(1) + sqrt(2)
= 1 + sqrt(2)