Difference between revisions of "2016 AMC 12A Problems/Problem 24"
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==Solution== | ==Solution== | ||
| + | <math>\text{The acceleration must be zero at the }x\text{-intercept: inflection point for minimum }a\text{ value to exist.}</math> | ||
| + | <math>x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}\text{ for inflection \& intercept}</math> | ||
| + | <math>\text{Minimum: }y=\left(x-\frac{a}{3}\right)^3</math> | ||
| + | <math>x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}</math> | ||
| + | <math>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</math> | ||
| + | <math>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{9}</math> | ||
| + | <math>f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}</math> | ||
| + | <math>f(x)=0\rightarrow x=\sqrt{3}\text{ triple root.}</math> | ||
| + | <math>\text{Complete the cube (!)}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:24, 5 February 2016
Problem
There is a smallest positive real number 
such that there exists a positive real number 
such that all the roots of the polynomial 
are real. In fact, for this value of the value of is unique. What is this value of ?
Solution
See Also
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 | |
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