Difference between revisions of "1965 AHSME Problems/Problem 22"
(created solution page) |
(→Solution) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 4: | Line 4: | ||
<math>a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )</math> holds: | <math>a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )</math> holds: | ||
| − | <math>\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 | + | <math>\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 \qquad |
\textbf{(B) }\ \text{for all values of }x \\ | \textbf{(B) }\ \text{for all values of }x \\ | ||
| − | \textbf{(C) }\ \text{only when }x = 0 | + | \textbf{(C) }\ \text{only when }x = 0 \qquad |
\textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\ | \textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\ | ||
| − | \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0 </math> | + | \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0 </math> |
== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | <math>a_0</math> cannot be <math>0</math>, because <math>a_2 \neq 0</math>, so the polynomial can take non-zero values when <math>a_0=0</math> and thereby not satisfy the equation. Expanding <math>a_0(1-\frac{x}{r})(1-\frac{x}{s})</math> gives us <math>a_0(1-\frac{x}{r}-\frac{x}{s}+\frac{x^2}{rs})</math>. Using [[Vieta's formulas]], we see that <math>rs=\frac{a_0}{a_2}</math>, so <math>a_0=a_2rs</math>. Plugging this into the expanded right hand side of the given equation, we see that that side equals <math>a_0-a_2(r+s)x+a_2x^2</math>. Using Vieta's formulas again, we equate this expression to <math>a_0+a_1x+a_2x^2</math>, which is the left hand side of the given equation. Thus, as long as <math>a_0 \neq 0</math>, the equation holds for all values of <math>x</math>. This fact corresponds with answer choice <math>\fbox{\textbf{(A)}}</math>. |
== See Also == | == See Also == | ||
Latest revision as of 17:50, 18 July 2024
Problem
If 
and 
and 
are the roots of 
, then the equality
holds:
Solution
cannot be 
, because , so the polynomial can take non-zero values when 
and thereby not satisfy the equation. Expanding 
gives us 
. Using Vieta's formulas, we see that 
, so 
. Plugging this into the expanded right hand side of the given equation, we see that that side equals 
. Using Vieta's formulas again, we equate this expression to 
, which is the left hand side of the given equation. Thus, as long as 
, the equation holds for all values of 
. This fact corresponds with answer choice 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
