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Difference between revisions of "2007 AMC 12A Problems/Problem 21"

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m (→‎Solution: typo)
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== Solution ==
 
== Solution ==
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By [[Vieta's formulas]], the sum of the roots of a [[quadratic equation]] is <math>\frac {-b}a</math>,  the product of the zeros is <math>\frac ca</math>, and the sum of the coefficients is <math>\displaystyle a + b + c</math>. Setting equal the first two tells us that <math>\frac {-b}{a} = \frac ca \Rightarrow b = -c</math>. Thus, <math>\displaystyle a + b + c = a + b - b = a</math>, so the common value is also equal to the coefficient of <math>x^2 \Longrightarrow \textrm{A}</math>.  
+
By [[Vieta's formulas]], the sum of the roots of a [[quadratic equation]] is <math>\frac {-b}a</math>,  the product of the zeros is <math>\frac ca</math>, and the sum of the coefficients is <math>a + b + c</math>. Setting equal the first two tells us that <math>\frac {-b}{a} = \frac ca \Rightarrow b = -c</math>. Thus, <math>a + b + c = a + b - b = a</math>, so the common value is also equal to the coefficient of <math>x^2 \Longrightarrow \textrm{A}</math>.  
  
 
To disprove the others, note that:
 
To disprove the others, note that:
 
*<math>\mathrm{B}</math>: then <math>b = \frac {-b}a</math>, which forces <math>a = -1</math> (not an [[identity]]).
 
*<math>\mathrm{B}</math>: then <math>b = \frac {-b}a</math>, which forces <math>a = -1</math> (not an [[identity]]).
−
*<math>\mathrm{C}</math>: the [[y-intercept]] is <math>c</math>, so <math>c - \frac ca</math> which forces <math>a = 1</math>.
+
*<math>\mathrm{C}</math>: the [[y-intercept]] is <math>c</math>, so <math>c = \frac ca</math> which forces <math>a = 1</math>.
 
*<math>\mathrm{D}</math>: an [[x-intercept]] of the graph is a root of the polynomial, but this excludes the other root.
 
*<math>\mathrm{D}</math>: an [[x-intercept]] of the graph is a root of the polynomial, but this excludes the other root.
 
*<math>\mathrm{E}</math>: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.
 
*<math>\mathrm{E}</math>: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.

Revision as of 19:57, 28 September 2007

Problem

The sum of the zeros, the product of the zeros, and the sum of the coefficients of the function $\displaystyle f(x)=ax^{2}+bx+c$ are equal. Their common value must also be which of the following?

$\textrm{(A)}\ \textrm{the\ coefficient\ of\ }x^{2}~~~ \textrm{(B)}\ \textrm{the\ coefficient\ of\ }x$ $\textrm{(C)}\ \textrm{the\ y-intercept\ of\ the\ graph\ of\ }y=f(x)$ $\textrm{(D)}\ \textrm{one\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$ $\textrm{(E)}\ \textrm{the\ mean\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$

Solution

By Vieta's formulas, the sum of the roots of a quadratic equation is $\frac {-b}a$, the product of the zeros is $\frac ca$, and the sum of the coefficients is $a + b + c$. Setting equal the first two tells us that $\frac {-b}{a} = \frac ca \Rightarrow b = -c$. Thus, $a + b + c = a + b - b = a$, so the common value is also equal to the coefficient of $x^2 \Longrightarrow \textrm{A}$.

To disprove the others, note that:

  • $\mathrm{B}$: then $b = \frac {-b}a$, which forces $a = -1$ (not an identity).
  • $\mathrm{C}$: the y-intercept is $c$, so $c = \frac ca$ which forces $a = 1$.
  • $\mathrm{D}$: an x-intercept of the graph is a root of the polynomial, but this excludes the other root.
  • $\mathrm{E}$: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.

See also

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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