1950 AHSME Problems/Problem 15

Problem

The real roots of $x^2+4$ are:

$\textbf{(A)}\ (x^{2}+2)(x^{2}+2)\qquad\textbf{(B)}\ (x^{2}+2)(x^{2}-2)\qquad\textbf{(C)}\ x^{2}(x^{2}+4)\qquad\\ \textbf{(D)}\ (x^{2}-2x+2)(x^{2}+2x+2)\qquad\textbf{(E)}\ \text{Non-existent}$

Solution

Solution 1

This looks similar to a difference of squares, so we can write it as $(x+2i)(x-2i).$ Neither of these factors are real.

Also, looking at the answer choices, there is no way multiplying two polynomials of degree $2$ will result in a polynomial of degree $2$ as well. Therefore the real factors are $\boxed{\mathrm{(E)}\text{ Non-existent.}}$

Solution 2

Let's try to find all real roots of $x^2+4=0$ . If there was a real root (call it $r$ ), then it would satisfy $r^2+4=0$ . Rearranging gives that $r^2=-4$ . Therefore a real root $r$ of $x^2+4=0$ would satisfy $r^2<0$ . However, the Trivial Inequality states that no real $r$ satisfies $r^2<0$ , which must mean that there are no real roots $r$ ; they are $\boxed{\mathrm{(E)}\text{ Non-existent.}}$

Solution 3 (Easier Solution)

Let us try to factor $x^2+4.$ We can see that the first term has variables only and the second has a constant only. Therefore, we cannot factor out anything and our answer is $\boxed{\mathrm{(E)}\text{ Non-existent.}}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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