1983 AIME Problems/Problem 3
If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with.
Instead, we substitute for , so that the equation becomes .
Now we can square; solving for , we get or . The second root is extraneous since is always non-negative (and moreover, plugging in , we get , which is obviously false). Hence we have as the only solution for . Substituting back in for ,
Both of the roots of this equation are real, since its discriminant is , which is positive. Thus by Vieta's formulas, the product of the real roots is simply .
We begin by noticing that the polynomial on the left is less than the polynomial under the radical sign. Thus: Letting , we have . Because the square root of a real number can't be negative, the only possible is .
Substituting that in, we have
Reasoning as in Solution 1, the product of the roots is .
Begin by completing the square on both sides of the equation, which gives Now by substituting , we get , or The solutions in are then Turns out, are a pair of extraneous solutions. Thus, our answer is then By difference of squares.
We are given the equation Squaring both sides yields Substituting yields Thus . However if , the left side of the equation would be negative while the right side is negative. Thus is the only possible value and we have Since the discriminant is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, .
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