AMC 8 2019 Problem 20
AMC8 2019 年第 20 题
方程与不等式★★★☆☆
How many different real numbers satisfy the equation
有多少个不同的实数 x 满足方程
- A.0
- B.1
- C.2
- D.4
- E.8
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
We have that if and only if . If , then , giving 2 solutions. If , then , giving 2 more solutions. All four of these solutions work, so the answer is 4. Further, the equation is a quartic in , so by the Fundamental Theorem of Algebra, there can be at most four real solutions.
We can expand to get , so now our equation is . Subtracting 16 from both sides gives us . Now, we can factor the left-hand side to get . If and/or equals 0, then the entire left side will equal 0. Since the solutions can be both positive and negative, we have 4 solutions: -3,3,-1,1 (we can find these solutions by setting and equal to 0 and solving for ). So, the answer is 4. Note: every positive root can also be shown as negative by WrenMath by Tarmac-boy
Subtract 16 from both sides and factor using difference of squares: (x+3)(x-3)(x+1)(x-1) = 0. Quite obviously, this equation has 4 solutions.
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