Question

Solve
5+2√6^2 + 5-2√6^2​

Answer 1

Answer:

The first step we will do is simplify the powers.

(5 + 2√6)² + (5–2√6)²

(5+2√6)(5+2√6) + (5–2√6)(5–2√6)

Then we multiply those parentheses together lets start with:

(5+2√6)(5+2√6)

There are two methods to multiplying binomials, I will show you the one I use:

We will muliply this going from first, outside, inside, and last. This can be remembered with the word FOIL:

(a + b) (a + b) The a’s are first because they are the most left. The a and b on the edges are the outside. The bold a and b are close together on inside. Finally, both b’s are the far right ao they are last. Using this:

(5+2√6)(5+2√6)

5² First

5² + 5*2√6 Outside

5² + 5*2√6 + 5*2√6 Inside

5² + 5*2√6 + 5*2√6 + (2√6)² Last

Repeating this with the other side:

5² + 5*2√6 + 5*2√6 + (2√6)² + 5² - 5*2√6 - 5*2√6 + (2√6)²

Note: notice the minuses on the second set. This is because of “–2√6” where it’s minus follows it, making it a negative number. Furthermore, see the last monomial (2√6)² is positive because:

-x * -x = x² so -(2√6) * -(2√6) = (2√6)²

We will now simplify this very long line.

5² + 5*2√6 + 5*2√6 + (2√6)² + 5² - 5*2√6 - 5*2√6 + (2√6)²

5² = 25 so lets change those

25 + 5*2√6 + 5*2√6 + (2√6)² + 25 - 5*2√6 - 5*2√6 + (2√6)²

Now let’s do the same for (2√6)²

(2√6)² = We can distribute the square to both to eliminate the parentheses:

2²√6², 2² = 4 so

4√6² The square of a root cancels eachother out so you are left with

4(6) = 24 Let’s put this back in

25 + 5*2√6 + 5*2√6 + 24 + 25 - 5*2√6 - 5*2√6 + 24

Let’s reorganize these:

25 + 25 + 24 + 24 + 5*2√6 + 5*2√6 - 5*2√6 - 5*2√6

Do you see how there is 5*2√6 and - 5*2√6 It’s the exact same monomial but negative, adding these will cancel them out.

25 + 25 + 24 + 24 + 5*2√6 - 5*2√6 Once again we have that same pair here so we cancel them out too.

25 + 25 + 24 + 24 We are left with these last four which added together will give you:

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