Question

Perform the following multiplication. The answer should be in simplified form for radical expressions.

Answer 1

\[(5\sqrt{a}+7\sqrt{b})(5\sqrt{a}-7\sqrt{b})\]

Answer 2

Is the problem the multiplication, or the simplification?

Answer 3

\[5\sqrt{a} \times 5\sqrt{a} +7\sqrt{b} \times 5\sqrt{a} - 7\sqrt{b}\times 5\sqrt{a} -7\sqrt{b}\times7\sqrt{b}\]

Answer 4

Actually, if you look at it for a moment, you can see that you don't need to do the multiplication at all, and can simply write down the answer!

Answer 5

true, but I think it important to make sure the process is known

Answer 6

25\[25a -49b\]

Answer 7

Difference of squares:
\[(a+b)(a-b) = a(a-b) + b(a-b) = a*a -a*b + a*b - b*b\]\[\qquad = a^2 -ab + ab - b^2\]\[\qquad=a^2\cancel{-ab}\cancel{+ab}-b^2\]\[\qquad=a^2-b^2\]

don't be confused by the fact that I used \(a\) and \(b\) and the problem uses them also. The concept is the same whether you use \(a\), \(z\), or anything in between.

Here you have \[(5\sqrt{a}+7\sqrt{b})(5\sqrt{a}-7\sqrt{b})\]That can be written using the difference of squares as
\[(5\sqrt{a})^2 -(7\sqrt{b})^2\]Do you know how to simplify those two terms?

Answer 8

25a-49b?

Answer 9

indeed!

\[5\sqrt{a}*5\sqrt{a} = 25\sqrt{a}\sqrt{a} = 25a\]\[7\sqrt{b}*7\sqrt{b} = 49b\]

So you should be able to look at that problem, say "aha, a difference of squares!" and immediately write down \[25a -49b\]as the answer.

Answer 10

You should already know how to multiply two binomials, if you want to do it the long way, but recognizing a difference of squares and going in either direction is the important lesson here, I believe.

Either direction meaning recognizing that \[(a-b)(a+b) = a^2-b^2\]OR that \[a^2-b^2 = (a-b)(a+b)\]

If I give you something like \[16x^2 -9y^2\]the little alarm bell should go off saying "hey, both terms are squares, and it's a difference, I can factor that as"\[(\sqrt{16x^2}+\sqrt{9y^2})(\sqrt{16x^2}-\sqrt{9y^2}) = (4x+3y)(4x-3y)\]

Answer 11

I need to remember that formula, ugh ! lol thank you !

Answer 12

yeah, that's like the single most useful factoring formula, and it comes back in all sorts of places...

I almost never think of it in these terms, but you can use it to do arithmetic, too:

\[7*11 = (9-2)(9+2) = 9^2-2^2 = 81-4 = 77\]

Answer 13

Think fondly of me every time you encounter it in the coming years, and I'll be soaking up all the good vibrations :-)