# 2019 AMC 8 Problems/Problem 20

## Problem 20

How many different real numbers $x$ satisfy the equation $$(x^{2}-5)^{2}=16?$$ $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$

## Solution 1

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.

## Solution 2

We can expand $(x^2-5)^2$ to get $x^4-10x^2+25$, so now our equation is $x^4-10x^2+25=16$. Subtracting $16$ from both sides gives us $x^4-10x^2+9=0$. Now, we can factor the left hand side to get $(x^2-9)(x^2-1)=0$. If $x^2-9$ and/or $x^2-1$ equals $0$, then the whole 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 $x^2-9$ and $x^2-1$ equal to $0$ and solving for $x$). So, the answer is $\boxed{\textbf{(D) }4}$.

~UnstoppableGoddess

## Solution 3

Subtract 16 from both sides and factor using difference of squares: $$(x^2 - 5)^2 = 16$$ $$(x^2 - 5)^2 - 16 =0$$ $$(x^2 - 5)^2 - 4^2 = 0$$ $$[(x^2 - 5)-4][(x^2 - 5) + 4] = 0$$ $$(x^2 - 9)(x^2 - 1) =0$$ $$(x+3)(x-3)(x+1)(x-1) = 0$$

Quite obviously, this equation has $\boxed{\textbf{(D) }4}$ solutions.

~TaeKim

## Solution 4

https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)

-Happytwin

## Video Solution

Solution detailing how to solve the problem: https://youtu.be/x4cF3o3Fzj8

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 