Difference between revisions of "1967 AHSME Problems/Problem 35"
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By Vieta, the sum of the roots is <math>-\frac{-144}{64} = \frac{9}{4}</math>. Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots. Therefore, <math>\frac{\frac{9}{4}}{3} = \frac{3}{4}</math> is the middle root. | By Vieta, the sum of the roots is <math>-\frac{-144}{64} = \frac{9}{4}</math>. Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots. Therefore, <math>\frac{\frac{9}{4}}{3} = \frac{3}{4}</math> is the middle root. | ||
| − | The other two roots have a sum of <math>\frac{9}{4} - \frac{3}{4} = \frac{3}{2}</math>. By Vieta on the original cubic, the product of all <math>3</math> roots is <math>-\frac{-15}{64} = \frac{15}{64}</math>, so the product of the remaining two roots is <math>\frac{\frac{15}{64}}{\frac{3}{4}} = \frac{5}{16}</math>. If the sum of the two remaining roots is <math>\frac{3}{2} = \frac{24}{16}</math>, and the product is <math>\frac{5}{16}</math>, the roots are also the two roots of <math>16x^2 - 24x + 5 = 0</math>. | + | The other two roots have a sum of <math>\frac{9}{4} - \frac{3}{4} = \frac{3}{2}</math>. By Vieta on the original cubic, the product of all <math>3</math> roots is <math>-\frac{-15}{64} = \frac{15}{64}</math>, so the product of the remaining two roots is <math>\frac{\frac{15}{64}}{\frac{3}{4}} = \frac{5}{16}</math>. If the sum of the two remaining roots is <math>\frac{3}{2} = \frac{24}{16}</math>, and the product is <math>\frac{5}{16}</math>, the two remaining roots are also the two roots of <math>16x^2 - 24x + 5 = 0</math>. |
| − | The two remaining roots are <math>x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}</math>, and they have a difference of <math>\frac{2\sqrt{B^2 - 4AC}{2A}</math>. Plugging in gives <math>\frac{\sqrt{24^2 - 4(16)(5)}{16}</math>, which is equal to <math>1</math>, which is answer <math>\fbox{B}</math>. | + | The two remaining roots are thus <math>x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}</math>, and they have a difference of <math>\frac{2\sqrt{B^2 - 4AC}}{2A}</math>. Plugging in gives <math>\frac{\sqrt{24^2 - 4(16)(5)}}{16}</math>, which is equal to <math>1</math>, which is answer <math>\fbox{B}</math>. |
== See also == | == See also == | ||
Revision as of 14:17, 12 July 2019
Problem
The roots of 
are in arithmetic progression. The difference between the largest and smallest roots is:
Solution
By Vieta, the sum of the roots is 
. Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots. Therefore, 
is the middle root.
The other two roots have a sum of 
. By Vieta on the original cubic, the product of all 
roots is 
, so the product of the remaining two roots is 
. If the sum of the two remaining roots is 
, and the product is 
, the two remaining roots are also the two roots of 
.
The two remaining roots are thus 
, and they have a difference of 
. Plugging in gives 
, which is equal to 
, which is answer 
.
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
