Difference between revisions of "1975 AHSME Problems/Problem 27"
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Megaboy6679 (talk | contribs) (→Solution 1) |
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<cmath>p^2 + q^2 + r^2 = -1.</cmath> | <cmath>p^2 + q^2 + r^2 = -1.</cmath> | ||
− | Therefore, <math>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}</math>. The answer is (E). | + | Therefore, <math>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}</math>. The answer is <math>\text(E)</math>. |
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==Solution 2(Faster)== | ==Solution 2(Faster)== | ||
We know that <math>p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr</math>. By Vieta's formulas, <math>p+q+r=1</math>,<math>pqr=2</math>, and <math>pq+qr+pr=1</math>. | We know that <math>p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr</math>. By Vieta's formulas, <math>p+q+r=1</math>,<math>pqr=2</math>, and <math>pq+qr+pr=1</math>. |
Revision as of 23:05, 18 March 2023
Problem
If and are distinct roots of , then equals
Solution 1
If is a root of , then , or Similarly, , and , so
By Vieta's formulas, , , and . Squaring the equation , we get Subtracting , we get
Therefore, . The answer is .
Solution 2(Faster)
We know that . By Vieta's formulas, ,, and . So if we can find , we are done. Notice that , so , which means that
~pfalcon