Question

what is equivalent to (11) / ((radical 3)-5)? can you explain how you did it?

Answer 1

\[\frac{ 11 }{ \sqrt{3}-5 }\]

Answer 2

thats what i mean ^

Answer 3

To simplify (write what's equivalent), you have mutliply both the top and bottom by the conjugate

For example if you had \[\Large \sqrt{a}+x\]
then the conjugate would be \[\Large \sqrt{a}-x\]

Answer 4

Therefore, we now have \[\Large \frac{11}{\sqrt{3}-5} \times \frac{\sqrt{3}+5}{\sqrt{3}+5}\]

Answer 5

ok i get that please continue im still a little confused

Answer 6

\[\Large \frac{11}{\sqrt{3}-5} \times \frac{\sqrt{3}+5}{\sqrt{3}+5}=\frac{11 \times (\sqrt{3}+5)}{\sqrt{3}^{2}-5^{2}}\]

Answer 7

\[\Large \frac{11\sqrt{3}+55}{3-25}\]
\[\Large \frac{11\sqrt{3}+55}{-22}\]

Answer 8

When I did the denominator: \[\Large \sqrt{3}-5 \times \sqrt{3}+5\]

I just remembered a rule that \[\Huge (a-b)^2=(a+b)(a-b)\]
So it follows that

Answer 9

\[\Large \sqrt{3}-5 \times \sqrt{3}+5=\sqrt{3}^2-5^2\]
where
a=sqrt(3)
b=5

Answer 10

ok i understand that but is there a way the simplify

\[\frac{ 11\sqrt{3}+55 }{ -22 }\]

any more?

Answer 11

Sorry the rule is \[\Huge a^2-b^2=(a+b)(a-b)\]

Answer 12

Yes, there is

Answer 13

Factor out 11 in the top and the bottom.

Answer 14

\[\frac{ 11\sqrt{3}+55 }{ -22 }=\frac{ 11*(\sqrt{3}+5 )}{11* (-2)}=-\frac{ \sqrt{3}+55 }{ 2 }\]

Answer 15

oh ok thank you so much