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Difference between revisions of "2019 AMC 12B Problems/Problem 21"
(Removed the second solution which was just the same as the first one; cleaned up the first solution's grammar and use of LaTeX, and added more detail for clarity)
 
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==Solution==
 
==Solution==
 +
Firstly, if <math>r=s</math>, then <math>a=b=c</math>, so the equation becomes <math>ax^2 + ax + a = 0 \Rightarrow x^2 + x + 1=0</math>, which has no real roots.
 +
 +
Hence there are three cases we need to consider:
 +
 +
'''Case 1''': <math>a=b=r</math> and <math>c=s \neq r</math>:
 +
The equation becomes <math>ax^2+ax+c=0</math>, and by Vieta's Formulas, we have <math>a+c=-1</math> and <math>ac = \frac{c}{a}</math>. This second equation becomes <math>(a^2-1)c=0</math>. Hence one possibility is <math>c=0</math>, in which case <math>a=-1</math>, giving the equation <math>-x^2 - x = 0</math>, which has roots <math>0</math> and <math>{-}1</math>. This gives one valid solution. On the other hand, if <math>c\neq 0</math>, then <math>a^2-1=0</math>, so <math>a = \pm 1</math>. If <math>a=1</math>, we have <math>c=-2</math>, and the equation is <math>x^2 + x - 2 = 0</math>, which clearly works, giving a second valid solution. If <math>a=-1</math>, then we have <math>c=0</math>, which has already been considered, so this possibility gives no further valid solutions.
 +
 +
'''Case 2''': <math>a=c=r</math>, <math>b=s \neq r</math>:
 +
The equation becomes <math>ax^2 + bx + a = 0</math>, so by Vieta's Formulas, we have <math>a+b = -\frac{b}{a}</math> and <math>ab = 1</math>. These equations reduce to <math>a^3 + a + 1 = 0</math>. By sketching a graph of this function, we see that there is exactly one real root. (Alternatively, note that as it is of odd degree, there is at least one real root, and by differentiation, it has no stationary points, so there is at most one real root. Combining these gives exactly one real root.) This gives a third valid solution.
 +
 +
'''Case 3''': <math>a=r</math>, <math>b=c=s \neq r</math>:
 +
The equation becomes <math>ax^2 + bx+b=0</math>, so by Vieta's Formulas, we have <math>a+b = -\frac{b}{a}</math> and <math>ab = \frac{b}{a}</math>.
 +
Observe that <math>b \neq 0</math>, as if it were <math>0</math>, the equation would just have one real root, <math>0</math>, so this would not give a valid solution. Thus, taking the second equation and dividing both sides by <math>b</math>, we deduce have <math>a=\frac{1}{a}</math>, so <math>a=\pm 1</math>. If <math>a=1</math>, we have <math>1+b=-b</math>, giving <math>b=-\frac{1}{2}</math>, so the equation is <math>x^2 - \frac{1}{2}x - \frac{1}{2} = 0</math>, which is a fourth valid solution. If <math>a=-1</math>, we have <math>1+b=b</math>, which is a contradiction, so this case gives no further valid solutions.
 +
 +
Hence the total number of valid solutions is <math>\boxed{\textbf{(B) } 4}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2019|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:16, 18 February 2019

Problem

How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bx+c,a\neq 0,$ and the roots are $r$ and $s,$ then the requirement is that $\{a,b,c\}=\{r,s\}$.)

$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}$

Solution

Firstly, if $r=s$, then $a=b=c$, so the equation becomes $ax^2 + ax + a = 0 \Rightarrow x^2 + x + 1=0$, which has no real roots.

Hence there are three cases we need to consider:

Case 1: $a=b=r$ and $c=s \neq r$: The equation becomes $ax^2+ax+c=0$, and by Vieta's Formulas, we have $a+c=-1$ and $ac = \frac{c}{a}$. This second equation becomes $(a^2-1)c=0$. Hence one possibility is $c=0$, in which case $a=-1$, giving the equation $-x^2 - x = 0$, which has roots $0$ and ${-}1$. This gives one valid solution. On the other hand, if $c\neq 0$, then $a^2-1=0$, so $a = \pm 1$. If $a=1$, we have $c=-2$, and the equation is $x^2 + x - 2 = 0$, which clearly works, giving a second valid solution. If $a=-1$, then we have $c=0$, which has already been considered, so this possibility gives no further valid solutions.

Case 2: $a=c=r$, $b=s \neq r$: The equation becomes $ax^2 + bx + a = 0$, so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = 1$. These equations reduce to $a^3 + a + 1 = 0$. By sketching a graph of this function, we see that there is exactly one real root. (Alternatively, note that as it is of odd degree, there is at least one real root, and by differentiation, it has no stationary points, so there is at most one real root. Combining these gives exactly one real root.) This gives a third valid solution.

Case 3: $a=r$, $b=c=s \neq r$: The equation becomes $ax^2 + bx+b=0$, so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = \frac{b}{a}$. Observe that $b \neq 0$, as if it were $0$, the equation would just have one real root, $0$, so this would not give a valid solution. Thus, taking the second equation and dividing both sides by $b$, we deduce have $a=\frac{1}{a}$, so $a=\pm 1$. If $a=1$, we have $1+b=-b$, giving $b=-\frac{1}{2}$, so the equation is $x^2 - \frac{1}{2}x - \frac{1}{2} = 0$, which is a fourth valid solution. If $a=-1$, we have $1+b=b$, which is a contradiction, so this case gives no further valid solutions.

Hence the total number of valid solutions is $\boxed{\textbf{(B) } 4}$.

See Also

2019 AMC 12B (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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