Difference between revisions of "1975 AHSME Problems/Problem 27"
Henry wang (talk | contribs) (Created page with "If <math>p</math> is a root of <math>x^3 - x^2 + x - 2 = 0</math>, then <math>p^3 - p^2 + p - 2 = 0</math>, or <cmath>p^3 = p^2 - p + 2.</cmath> Similarly, <math>q^3 = q^2 - q...") |
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+ | ==Problem== | ||
+ | If <math>p, q</math> and <math>r</math> are distinct roots of <math>x^3-x^2+x-2=0</math>, then <math>p^3+q^3+r^3</math> equals | ||
+ | |||
+ | <math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}</math> | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 1== | ||
If <math>p</math> is a root of <math>x^3 - x^2 + x - 2 = 0</math>, then <math>p^3 - p^2 + p - 2 = 0</math>, or | If <math>p</math> is a root of <math>x^3 - x^2 + x - 2 = 0</math>, then <math>p^3 - p^2 + p - 2 = 0</math>, or | ||
<cmath>p^3 = p^2 - p + 2.</cmath> | <cmath>p^3 = p^2 - p + 2.</cmath> | ||
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<cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath> | <cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath> | ||
− | By Vieta's formulas, <math>p + q + r = 1</math>, <math>pq + pr + qr = 1</math>, and <math>pqr = 2</math>. Squaring the equation <math>p + q + r = 1</math>, we get | + | By [[Vieta's formulas]], <math>p + q + r = 1</math>, <math>pq + pr + qr = 1</math>, and <math>pqr = 2</math>. Squaring the equation <math>p + q + r = 1</math>, we get |
<cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath> | <cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath> | ||
Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get | Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get | ||
<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)== | ||
+ | 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>. | ||
+ | So if we can find <math>p^2+q^2+r^2</math>, we are done. Notice that <math>(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr</math>, so <math>p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1</math>, which means that <math>p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{\text{(E) }4}</math> | ||
+ | |||
+ | ~pfalcon | ||
+ | |||
+ | ==Solution 3 (Beginner's Solution)== | ||
+ | Use Vieta's formulas to get <math>p+q+r=1</math>, <math>pq+qr+pr=1</math>, and <math>pqr=2</math>. | ||
+ | |||
+ | Square <math>p+q+r=1</math>, and get <math>p^2+q^2+r^2+2pq+2pr+2qr=1</math> | ||
+ | |||
+ | Substitute <math>pq+qr+pr=1</math> and simplify to get <math>-1=p^2+q^2+r^2</math> | ||
+ | |||
+ | After that, multiply both sides by <math>1=p+q+r</math>, to get | ||
+ | <math>-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q</math> | ||
+ | |||
+ | Then, factor out <math>pq</math>, <math>qr</math>, and <math>pr</math>: | ||
+ | <math>-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)</math> | ||
+ | |||
+ | Then, substitute the first equation into <math>p+q</math>, <math>q+r</math>, and <math>p+r</math>. | ||
+ | <math>-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)</math> | ||
+ | |||
+ | Then, multiply it out: | ||
+ | <math>-1=p^3+q^3+r^3+pq+qr+pr-3pqr</math> | ||
+ | |||
+ | After that, substitute the equations <math>pq+qr+pr=1</math> and <math>pqr=2</math>: | ||
+ | <math>-1=p^3+q^3+r^3+1-6</math> | ||
+ | |||
+ | Solving that, you get <math>p^3+q^3+r^3=\boxed{\text{(E) }4}</math> | ||
+ | ~EZ PZ Ms.Lemon SQUEEZY |
Latest revision as of 18:58, 23 August 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
Solution 3 (Beginner's Solution)
Use Vieta's formulas to get ,
, and
.
Square , and get
Substitute and simplify to get
After that, multiply both sides by , to get
Then, factor out ,
, and
:
Then, substitute the first equation into ,
, and
.
Then, multiply it out:
After that, substitute the equations and
:
Solving that, you get
~EZ PZ Ms.Lemon SQUEEZY