Difference between revisions of "1975 AHSME Problems/Problem 27"
Megaboy6679 (talk | contribs) (→Solution 2(Faster)) |
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>. |
+ | |||
==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 00:05, 19 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