Difference between revisions of "1976 AHSME Problems/Problem 27"
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Since <math>y>0,</math> we have <math>y=\sqrt{2}-1.</math> | Since <math>y>0,</math> we have <math>y=\sqrt{2}-1.</math> | ||
− | Finally, the answer is < | + | Finally, the answer is <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath> |
~Someonenumber011 (Fundamental Logic) | ~Someonenumber011 (Fundamental Logic) | ||
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On the other hand, we have <cmath>y=\sqrt2-1</cmath> by the argument of either Solution 1 or Solution 2. | On the other hand, we have <cmath>y=\sqrt2-1</cmath> by the argument of either Solution 1 or Solution 2. | ||
− | Finally, the answer is < | + | Finally, the answer is <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Latest revision as of 12:41, 8 September 2021
Contents
Problem
If then equals
Solution 1
Let and
Note that Since we have
On the other hand, note that Since we have
Finally, the answer is
~Someonenumber011 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Let and
Note that We rewrite each term in the numerator separately:
- Let for some nonnegative rational numbers and We square both sides of this equation, then rearrange:
It follows that
By inspection, we get Alternatively, we conclude that and are the solutions to the quadratic equation by Vieta's Formulas, in which
Therefore, we obtain
- Similarly, we obtain
Substituting these results into we have On the other hand, we have by the argument of either Solution 1 or Solution 2.
Finally, the answer is
~MRENTHUSIASM
See also
1976 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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