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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<math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}</math>
 
<math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}</math>
  
==Solution 1==
+
==Solution==
First, if <math>r=s</math>, then <math>a=b=c</math>, equation <math>ax^2 + ax + a = 0</math> has no real roots.
+
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.
  
Then there are three cases:
+
Hence there are three cases we need to consider:
  
Case 1: <math>a=b=r</math>, <math>c=s</math>:
+
'''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.
  
The equation becomes <math>ax^2+ax+c=0</math>, and by Vieta's theorem, we have <math>a+c=-1</math>, <math>ac = \frac{c}{a}</math> or <math>(a^2-1)c=0</math>
+
'''Case 2''': <math>a=c=r</math>, <math>b=s \neq r</math>:
if <math>c=0</math>, we get <math>a=-1</math>, equation <math>-x^2 - x = 0</math> has roots 0 and <math>-1</math>. <math>\boxed{1}</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.
  
if <math>c\ne 0</math>, then <math>a^2-1=0</math>,  
+
'''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.
  
if <math>a=1</math>, we have <math>c=-2</math>, and the equation is <math>x^2 + x - 2 = 0</math> <math>\boxed{2}</math>
+
Hence the total number of valid solutions is <math>\boxed{\textbf{(B) } 4}</math>.
 
 
if <math>a=-1</math>, we have <math>c=0</math>, same as <math>\boxed{1}</math>
 
 
 
Case 2: <math>a=c=r</math>, <math>b=s</math>:
 
 
 
The equation becomes <math>ax^2 + bx + a = 0</math>, by Vieta's theorem, we have <math>a+b = -\frac{b}{a}</math> and <math>ab = 1</math>. This can be rewritten as
 
<math>a^3 + a + 1 = 0</math>, which has 1 real root of <math>a</math>. <math>\boxed{3}</math>
 
 
 
Case 3: <math>a=r</math>, <math>b=c=s</math>:
 
 
 
The equation becomes <math>a^x + bx+b=0</math>, by Vieta's theorem, we have <math>a+b = -\frac{b}{a}</math> and <math>ab = \frac{b}{a}</math>.
 
<math>b</math> can not be 0, so we have <math>a=\frac{1}{a}</math>, or <math>a=\pm 1</math>.
 
 
 
if <math>a=1</math>, we have <math>1+b=-b</math>, and <math>b=-\frac{1}{2}</math>, the equation is <math>x^2 - \frac{1}{2}x - \frac{1}{2} = 0</math>, <math>\boxed{4}</math>
 
 
 
if <math>a=-1</math>, we have <math>1+b=b</math>, a contradiction.
 
 
 
The answer is <math>\boxed{B}</math>
 
 
 
==Solution 2==
 
 
 
Because there are three coefficients and two roots, we need at least two elements in the set <math>\{a,b,c\}</math> to be equal to each other. It is possible that all three could be equal to each other. In the case that two elements in the set <math>\{a,b,c\}</math> are equal to each other, two of those elements will be equal to <math>r</math> and the third will be equal to <math>s</math>.
 
 
 
Case 1: <math>r = s</math>
 
 
 
We would need the polynomial <math>rx^2 + rx + r</math> to have a double root <math>r</math>. By inspection, there is no such polynomial, so there are no polynomials for this case.
 
 
 
Case 2: <math>a = b = r</math> and <math>c = s</math>
 
 
 
The polynomial will be in the form <math>rx^2 + rx + s</math>. By Vieta's formulas, <math>r + s = -1</math> and <math>rs = \frac{s}{r}</math>. The second equation tells us that either <math>r = 0</math> or <math>s^2 = 1</math>. Testing each possibility, we find the polynomials <math>-x^2 - x</math> and <math>x^2 + x - 2</math>, both of which work. There are 2 polynomials for this case.
 
 
 
Case 3: <math>a = c = r</math> and <math>b = s</math>
 
 
 
The polynomial will be in the form <math>rx^2 + sx + r</math>. By Vieta's formulas, <math>r + s = \frac{-s}{r}</math> and <math>rs = 1</math>. Through substitution, we get <math>r^3 + r + 1</math>. The function <math>f(r) = r^3 + r + 1</math> is a strictly increase function with one real root. This real root works for our value of <math>r</math>, so there is 1 polynomial in this case.
 
 
 
Case 4: <math>b = c = r</math> and <math>a = s</math>
 
 
 
The polynomial will be in the form <math>sx^2 + rx + r</math>. By Vieta's formulas, <math>r + s = \frac{-r}{s}</math> and <math>rs = \frac{r}{s}</math>. The second equation tells us that either <math>r = 0</math> or <math>s^2 = 1</math>. If <math>r = 0</math>, <math>s = 0</math>, which causes a divide by 0 error. If <math>s = -1</math>, the first equation becomes <math>0 = 1</math>, a contradiction. If <math>s = 1</math> then <math>r = \frac{-1}{2}</math>. There is 1 polynomial in this case.
 
 
 
Our answer is <math>\boxed{\mathbf{(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 00:16, 19 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 (ProblemsAnswer KeyResources)
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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