Question

What is the simplest form of the expression? See image attatched.
Can someone show the work of how to get to the answer. I know nothing about algebra Thanks!

Answer 1

I strongly suspect the answer is the last one worth 1 point because all the others are worth 0 points. :D Let's do the math to be sure.

What do we know about square roots?
Check your book or search engine for the Product Property of Square Roots. You should see that
\[\sqrt{a \times b}=\sqrt{a} \times \sqrt{b}\]
for 8 that would look like
\[\sqrt{8}=\sqrt{(2)(4)}=(\sqrt{2})(\sqrt{4})=2\sqrt{2}\]

With that in mind can we factor any of the square roots to lower terms?

Answer 2

the 6?

Answer 3

for real I know nothing. Sorry

Answer 4

That's why this is so much fun. You get to learn. It takes some hard work and reading but it is worth it.

So yeah, 6 is 2x3 so
\[\sqrt{6}=\sqrt{2}\sqrt{3}\]
If you put that in the given equations is there anything we can factor out?
\[\frac{ \sqrt{2}-\sqrt{2}\sqrt{3} }{ \sqrt{2}+\sqrt{2}\sqrt{3} }\]

Answer 5

oh, I think I figured it out hopefully lol I have to move on to the next quiz but Thank you for your help I think I get it a little better now so thank you a lot!

Answer 6

Okay, but there is one more part you have to do. Hopefully you will come back and see this later.

\[\frac{ 1-\sqrt{3} }{ 1+\sqrt{3}}= \frac{ 1 }{ 1 } \times \frac{ 1-\sqrt{3} }{ 1+\sqrt{3} } = \frac{ 1-\sqrt{3} }{ 1-\sqrt{3}} \times \frac{ 1-\sqrt{3} }{ 1+\sqrt{3}}\]
We ca do that because something divided by itself is equal to 1 and multiplying anything by one yields a product equal to itself.

If we do the multiplication we get
\[\frac{ 1^2-\sqrt{3}-\sqrt{3}+\sqrt{3}^2 }{ 1^2-\sqrt{3}+\sqrt{3}-\sqrt{3}^2 }\]

That equals
\[\frac{ 1-2\sqrt{3}+3 }{ 1-3 }=\frac{ 4-2\sqrt{3} }{ -2 }=-2+\sqrt{3}\]
which means that the (1pt) was a hint for us.