Difference between revisions of "2022 AMC 12A Problems/Problem 15"
Phuang1024 (talk | contribs) (Created page with "==Problem== The roots of the polynomial <math>10x^3 - 39x^2 + 29x - 6</math> are the height, length, and width of a rectangular box (right rectangular prism. A new rectangu...") |
Sugar rush (talk | contribs) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | The roots of the polynomial | + | The roots of the polynomial <math>10x^3 - 39x^2 + 29x - 6</math> are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 2 |
− | |||
− | <math>10x^3 - 39x^2 + 29x - 6</math> | ||
− | |||
− | are the height, length, and width of a rectangular box (right rectangular prism. A | ||
− | new rectangular box is formed by lengthening each edge of the original box by 2 | ||
units. What is the volume of the new box? | units. What is the volume of the new box? | ||
Line 24: | Line 19: | ||
- phuang1024 | - phuang1024 | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC12 box|year=2022|ab=A|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Revision as of 19:47, 11 November 2022
Problem
The roots of the polynomial are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 2 units. What is the volume of the new box?
Solution
Let , , be the three roots of the polynomial. The lenghtened prism's area is .
By vieta's formulas, we know that:
.
We can substitute these into the expression, obtaining
- phuang1024
See also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.