How can I solve 3/1+root 5 ?
@Nicoley269 would you mind taking a screenshot of the problem you're working on and upload it here?
\[\frac{ 3 }{ 1+\sqrt{5} }\]
^what u posted is an expression is the question a rationalization question?
multiply numerator and denominator by the conjugate of 1+sqrt{5}
What @Nnesha wrote is the the correct problem.
I don’t know how
If that's the case, multiply top and bottom by \(1 - \sqrt{5}\)
do you understand the purpose of that step
I got up to\[3-3\sqrt{5} /1+\sqrt{5} *1-\sqrt{5}\]
Great! So far you have \(\dfrac{3 - 3\sqrt{5}}{(1 + \sqrt{5})(1 - \sqrt{5})}\) All you have to do is just multiply the binomials in the denominator to do that you use the distributive property. Hopefully you're familiar with it: \(a(b + c) = ab + ac\) In this case, let \(a\) represent \(1 + \sqrt{5}\) \(b\) represent \(1\) \(c\) represent \(-\sqrt{5}\)
I can show you an example of how to multiply two binomials if you like
I I don’t get it. Would the denominator be 2?
yes please
Here's an example: \(\color\green{a}{(\color\red{b} + \color\purple{c})} = \color\green{a}\color\red{b} + \color\green{a}\color\purple{c}\) \( (\color\green{2 + \sqrt{6}})(\color\red{2} - \color\purple{\sqrt{6}}) \\= \color\red{2}\color\green{(2 + \sqrt{6})} \color\purple{- \sqrt{6}}\color\green{(2 + \sqrt{6})} \\=4 + 2\sqrt{6} - 2\sqrt{6} - 6 \\=4 - 6 \\=-2 \)
Hope that's not too confusing for you
oh. so i was correct. 2
That was just an example. I'd like to see you try multiplying the denominator on your own and posting all your steps here.
\[1(1-\sqrt{5})-\sqrt{5}(1+\sqrt{5})\]
Mistakes in red: \(1(1\color\red{-}\sqrt{5})-\sqrt{5}(1+\sqrt{5})\)
i did minus.
The binomials in parentheses must be the same in order to multiply correctly
ok...
i got 2
Please show your work. All your steps.
nvm. its -4
|dw:1526178979452:dw| actually its 6
oops
6 is correct
|dw:1526179472176:dw|
thank you!
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