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Solve: 5 over quantity of 4 plus square root of 5
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Multiply numerator and denominator by the conjugate of 4+sqrt5
\[\frac{5}{(4+\sqrt{5})}\times\frac{(4-\sqrt{5})}{(4-\sqrt{5})}\]
Can you finish that?
thats a viable option, but how is that a solution to an expression?
so its 20 - sqrt of 5 over 11?
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You getting warm. Multiply both terms of the binomial by 5 in the numerator
\[5(4-\sqrt{5})=20-5\sqrt{5}\]
and then that over 11?
Yes.
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