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1963 TMTA High School Algebra I Contest Problem 25

Problem

If $\sqrt{5}=2.236,$ the value of $\frac{1+\sqrt{5}}{3-2\sqrt{5}}$ to the nearest hundredth is:

$\text{(A)} \quad -2.25 \quad \text{(B)} \quad -2.35 \quad \text{(C)} \quad -2.15 \quad \text{(D)} \quad -2.20 \quad \text{(E)} \quad -2.50$

Solution

Simplify the expression first: \[\frac{1+\sqrt{5}}{3-2\sqrt{5}} \cdot \frac{3+2\sqrt{5}}{3+2\sqrt{5}}=\frac{3+3\sqrt{5}+2\sqrt{5}+10}{-17}=\frac{-13+5\sqrt{5}}{11}\] Then we can use our approximation $\sqrt{5}=2.236.$ We see that this is closest to $\boxed{\text{(C)} \quad -2.15}.$

See Also

1963 TMTA High School Mathematics Contests (Problems)
Preceded by
Problem 24 		TMTA High School Mathematics Contest Past Problems/Solutions Followed by
Problem 26
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