View source for 1963 TMTA High School Algebra I Contest Problem 25

== Problem ==
If <math>\sqrt{5}=2.236,</math> the value of <math>\frac{1+\sqrt{5}}{3-2\sqrt{5}}</math> to the nearest hundredth is:

<math>\text{(A)} \quad -2.25 \quad \text{(B)} \quad -2.35 \quad \text{(C)} \quad -2.15 \quad \text{(D)} \quad -2.20 \quad \text{(E)} \quad -2.50</math>

== Solution ==
Simplify the expression first:
<cmath>\frac{1+\sqrt{5}}{3-2\sqrt{5}} \cdot \frac{3+2\sqrt{5}}{3+2\sqrt{5}}=\frac{3+3\sqrt{5}+2\sqrt{5}+10}{-17}=\frac{-13+5\sqrt{5}}{11}</cmath>
Then we can use our approximation <math>\sqrt{5}=2.236.</math> We see that this is closest to <math>\boxed{\text{(C)} \quad -2.15}.</math>
== See Also ==
{{Succession box
|header=[[1963 TMTA High School Mathematics Contests]] ([[1963 TMTA High School Algebra I Contest Problems|Problems]])
|before=[[1963 TMTA High School Algebra I Contest Problem 24| Problem 24]]
|title=[[TMTA High School Mathematics Contest Past Problems/Solutions]]
|after=[[1963 TMTA High School Algebra I Contest Problem 26|Problem 26]]
}}