1984 USAMO Problems/Problem 1
In the polynomial , the product of of its roots is . Find .
Solution 1 (ingenious)
Using Vieta's formulas, we have:
From the last of these equations, we see that . Thus, the second equation becomes , and so . The key insight is now to factor the left-hand side as a product of two binomials: , so that we now only need to determine and rather than all four of .
Let and . Plugging our known values for and into the third Vieta equation, , we have . Moreover, the first Vieta equation, , gives . Thus we have two linear equations in and , which we solve to obtain and .
Therefore, we have , yielding .
Solution 2 (cool)
We start as before: and . We now observe that a and b must be the roots of a quadratic, , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic .
Equating the coefficients of and with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of and get
Let the roots of the equation be and . By Vieta's, Since and , then, . Notice thatcan be factored intoFrom the first equation, . Substituting it back into the equation,Expanding,So, and . Notice thatPlugging all our values in,
Video Solution by Omega Learn
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