Question

Square root of 4/3 minus square root of 3/4. How do you solve this?

Answer 1

\[\sqrt{\frac{ 4 }{ 3 }} - \sqrt{\frac{ 3 }{ 4 }}\]

Answer 2

let's do some simplifying first

Answer 3

we can actually split it up into two separate radicals like this

\[\sqrt{\frac{ a }{ b }} = \frac{ \sqrt{a} }{ \sqrt{b} }\]

\[\frac{ \sqrt{4} = 2 }{ \sqrt{3} }-\frac{ \sqrt{3} }{ \sqrt{4} =2}\]


\[\frac{ 2 }{ \sqrt{3} }-\frac{ \sqrt{3} }{ 2}\]

Answer 4

@Nancyg94 do you follow?

Answer 5

Yes, I follow!

Answer 6

I am stuck after this. I was thinking I could make the denominator the same. But that doesn't give me the answer I am looking for.

Answer 7

Ok I tried again! I've got it now!

Answer 8

Thank you!!

Answer 9

Is there anyway you can help me factor a quadratic? This is the only other problem I have had trouble with. Factor: x^5+2x^3-x^2-2

Answer 10

\[\sqrt{\frac{ 4 }{ 3 }}-\sqrt{\frac{ 3 }{ 4 }}=\frac{ 2 }{ \sqrt{3} }-\frac{ \sqrt{3} }{ 2 }=\frac{ 2 }{ \sqrt{3} }\times \frac{ \sqrt{3} }{ \sqrt{3}}-\frac{ \sqrt{3} }{ 2 }\]
\[=\frac{ 2\sqrt{3} }{ 3 }-\frac{ \sqrt{3} }{ 2 }=\left( \frac{ 4\sqrt{3}-3\sqrt{3} }{ 6 } \right)=\frac{ \sqrt{3} }{ 6 }\]
or\[\frac{ 1 }{ 2\sqrt{3} }\]

Answer 11

To start with take x^3 common from first two and -1 from last two

Answer 12

((X^3)-1) (x^2 +2x) ((x^2)+2)

Answer 13

\[x^3\left( x^2+2 \right)-1\left( x^2+2 \right)=\left( x^2+2 \right)\left( x^3-1 \right)\]
\[=\left( x^2+2 \right)\left( x-1 \right)\left( x^2+x+1 \right)\]
\[a^3-b^3=\left( a-b \right)\left( a^2+ab+b^2 \right)\]

Answer 14

\[\frac{ 2 }{ \sqrt{3} }-\frac{ \sqrt{3} }{ 2 } = \frac{ 4-3 }{ 2\sqrt{3} } = \frac{ 1 }{ 2\sqrt{3} }\]

Answer 15

I see where I made my mistake! Thank you!

Answer 16

That was a real big help!

Answer 17

a faster way to do it would be this.