Answer: [tex]2\sqrt{2}+\sqrt{3}\\\\[/tex]
a = 2 and b = 1
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Explanation:
Set the expression equal to the given form we want. Then square both sides so we get rid of the outer-most square root
[tex]\sqrt{11+4\sqrt{6}} = a\sqrt{2}+b\sqrt{3}\\\\\left(\sqrt{11+4\sqrt{6}}\right)^2 = \left(a\sqrt{2}+b\sqrt{3}\right)^2\\\\11+4\sqrt{6} = \left(a\sqrt{2}\right)^2+2*a\sqrt{2}*b\sqrt{3}+\left(b\sqrt{3}\right)^2\\\\11+4\sqrt{6} = 2a^2+2ab\sqrt{2*3}+3b^2\\\\11+4\sqrt{6} = 2a^2+3b^2+2ab\sqrt{6}\\\\[/tex]
In the third line, I used the rule that (x+y)^2 = x^2+2xy+y^2
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At this point, we equate the non-radical and radical terms to get this system of equations
[tex]\begin{cases}11 = 2a^2+3b^2\\ 4\sqrt{6} = 2ab\sqrt{6}\end{cases}[/tex]
The second equation turns into 4 = 2ab when we divide both sides by sqrt(6)
Then 4 = 2ab turns into ab = 2 after dividing both sides by 2.
We're told that a,b are rational numbers. Let's assume that they are integers (which is a subset of the rational numbers).
If so, then we have these four possibilities
- a = 2, b = 1
- a = -2, b = -1
- a = 1, b = 2
- a = -1, b = -2
If a,b are negative, then you'll find that [tex]a\sqrt{2}+b\sqrt{3}[/tex] overall is negative. But this contradicts that [tex]\sqrt{11+4\sqrt{6}}[/tex] is positive. So a,b must be positive.
Let's assume that a = 1 and b = 2. If so, then,
2a^2+3b^2 = 2(1)^2+3(2)^2 = 14
but we want that result to be 11 instead.
Let's try a = 2 and b = 1
2a^2+3b^2 = 2(2)^2+3(1)^2 = 11
which works out perfectly.
Therefore,
[tex]\sqrt{11+4\sqrt{6}} = 2\sqrt{2}+\sqrt{3}\\\\[/tex]
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Checking the answer:
Use a calculator to find that
[tex]\sqrt{11+4\sqrt{6}} \approx 4.5604779\\\\2\sqrt{2}+\sqrt{3} \approx 4.5604779\\\\[/tex]
both have the same decimal approximation, so this is a fairly informal way to confirm the answer.
Another thing you can do is to take advantage of the idea that if x = y, then x-y = 0
So if you want to see if two things are equal, you subtract them. You should get exactly 0 or something very small (pretty much equal to 0).
