Answer: [tex]\sqrt{3}+\sqrt{5}[/tex]
On a keyboard, we could type this as sqrt(3) + sqrt(5)
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Explanation:
The given expression is considered a nested radical because one radical, the [tex]\sqrt{15}[/tex], is buried inside another radical.
Let's assume that nested radical is of the form [tex]\sqrt{a}+\sqrt{b}[/tex] where a & b are positive real numbers.
Set the given expression equal to [tex]\sqrt{a}+\sqrt{b}[/tex] and square both sides to see what happens. We'll do a bit of algebra to simplify and rearrange things a bit as well. See the steps below.
[tex]\sqrt{8+2\sqrt{15}} = \sqrt{a}+\sqrt{b}\\\\\left(\sqrt{8+2\sqrt{15}}\right)^2 = \left(\sqrt{a}+\sqrt{b}\right)^2\\\\8+2\sqrt{15} = \left(\sqrt{a}\right)^2 + 2\sqrt{a}\sqrt{b} + \left(\sqrt{b}\right)^2\\\\8+2\sqrt{15} = a + 2\sqrt{ab} + b\\\\8+2\sqrt{15} = (a+b) + 2\sqrt{ab}\\\\[/tex]
If we equate terms, we get this system of equations
[tex]\begin{cases}8 = a+b\\2\sqrt{15} = 2\sqrt{ab}\\\end{cases}[/tex]
Solve the first equation for 'a' to get a = 8-b
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Let's plug a = 8-b into the second equation and solve for 'b'
[tex]2\sqrt{15} = 2\sqrt{ab}\\\\\sqrt{15} = \sqrt{ab}\\\\15 = ab\\\\15 = (8-b)b \ \ \text{ .... replace a with 8-b}\\\\15 = 8b-b^2\\\\b^2-8b+15 = 0\\\\(b-3)(b-5) = 0\\\\b-3 = 0 \ \text{ or } \ b-5 = 0\\\\b = 3 \ \text{ or } \ b = 5\\\\[/tex]
If b = 3, then a = 8-b = 8-3 = 5
If b = 5, then a = 8-b = 8-5 = 3
We have this symmetry going on with 'a' and b. If one value is 3, then the other is 5, and vice versa. The order doesn't matter.
That means the equation
[tex]\sqrt{8+2\sqrt{15}} = \sqrt{a}+\sqrt{b}\\\\[/tex]
updates to
[tex]\sqrt{8+2\sqrt{15}} = \sqrt{3}+\sqrt{5}\\\\[/tex]
The order doesn't matter on the right side since we can add two numbers in any order.
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You can use your calculator to confirm the answer.
Note that
[tex]\sqrt{8+2\sqrt{15}} \approx 3.968118785[/tex]
and
[tex]\sqrt{3}+\sqrt{5} \approx 3.968118785[/tex]
both result in the same decimal approximation to help show the two sides are equal.
