Question

Can someone help i don't get this problem

Answer 1

Answer:   $\sqrt{3}+\sqrt{5}$

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 $\sqrt{15}$, is buried inside another radical.

Let's assume that nested radical is of the form $\sqrt{a}+\sqrt{b}$ where a & b are positive real numbers.

Set the given expression equal to $\sqrt{a}+\sqrt{b}$ 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.

$\sqrt{8+2\sqrt{15}} = \sqrt{a}+\sqrt{b}\\\\\left(\sqrt{8+2\sqrt{15}}\right)^2 = \left(\sqrt{a}+\sqrt{b}\right)^2\\\\8+2\sqrt{15} = \left(\sqrt{a}\right)^2 + 2\sqrt{a}\sqrt{b} + \left(\sqrt{b}\right)^2\\\\8+2\sqrt{15} = a + 2\sqrt{ab} + b\\\\8+2\sqrt{15} = (a+b) + 2\sqrt{ab}$

If we equate terms, we get this system of equations

$\begin{cases}8 = a+b\\2\sqrt{15} = 2\sqrt{ab}\\\end{cases}$

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'

$2\sqrt{15} = 2\sqrt{ab}\\\\\sqrt{15} = \sqrt{ab}\\\\15 = ab\\\\15 = (8-b)b \ \ \text{ .... replace a with 8-b}\\\\15 = 8b-b^2\\\\b^2-8b+15 = 0\\\\(b-3)(b-5) = 0\\\\b-3 = 0 \ \text{ or } \ b-5 = 0\\\\b = 3 \ \text{ or } \ b = 5$

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

$\sqrt{8+2\sqrt{15}} = \sqrt{a}+\sqrt{b}$

updates to

$\sqrt{8+2\sqrt{15}} = \sqrt{3}+\sqrt{5}$

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

$\sqrt{8+2\sqrt{15}} \approx 3.968118785$

and

$\sqrt{3}+\sqrt{5} \approx 3.968118785$

both result in the same decimal approximation to help show the two sides are equal.

Answer 2

Answer:

the answer is 10•57(approximated as 10•6)