2019 AMC 12B Problems/Problem 21
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 and the roots are and then the requirement is that .)
First, if , then , equation has no real roots.
Then there are three cases:
Case 1: , :
The equation becomes , and by Vieta's theorem, we have , or if , we get , equation has roots 0 and .
if , then ,
if , we have , and the equation is
if , we have , same as
Case 2: , :
The equation becomes , by Vieta's theorem, we have and . This can be rewritten as , which has 1 real root of .
Case 3: , :
The equation becomes , by Vieta's theorem, we have and . can not be 0, so we have , or .
if , we have , and , the equation is ,
if , we have , a contradiction.
The answer is
Because there are three coefficients and two roots, we need at least two elements in the set 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 are equal to each other, two of those elements will be equal to and the third will be equal to .
We would need the polynomial to have a double root . By inspection, there is no such polynomial, so there are no polynomials for this case.
Case 2: and
The polynomial will be in the form . By Vieta's formulas, and . The second equation tells us that either or . Testing each possibility, we find the polynomials and , both of which work. There are 2 polynomials for this case.
Case 3: and
The polynomial will be in the form . By Vieta's formulas, and . Through substitution, we get . The function is a strictly increase function with one real root. This real root works for our value of , so there is 1 polynomial in this case.
Case 4: and
The polynomial will be in the form . By Vieta's formulas, and . The second equation tells us that either or . If , , which causes a divide by 0 error. If , the first equation becomes , a contradiction. If then . There is 1 polynomial in this case.
Our answer is
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