AMC8 2019

AMC8 2019 · Q20

AMC8 2019 · Q20. It mainly tests Absolute value, Quadratic equations.

How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]

方程 \[(x^{2}-5)^{2}=16?\] 有多少个不同的实数解 x？

(A) 0 0

(B) 1 1

(D) 4 4

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$. If $x^2-5 = 4$, then $x^2 = 9 \implies x = \pm 3$, giving 2 solutions. If $x^2-5 = -4$, then $x^2 = 1 \implies x = \pm 1$, giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{\textbf{(D) }4}$. Further, the equation is a quartic in $x$, so by the Fundamental Theorem of Algebra, there can be at most four real solutions.

(x^2-5)^2 = 16 当且仅当 x^2-5 = \pm 4。如果 x^2-5 = 4，则 x^2 = 9 \implies x = \pm 3，得 2 个解。如果 x^2-5 = -4，则 x^2 = 1 \implies x = \pm 1，得另外 2 个解。四个解都有效，所以答案是 \boxed{\textbf{(D) }4}。此外，该方程是 x 的四次方程，根据代数基本定理，最多有四个实根。

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