Difference between revisions of "1950 AHSME Problems/Problem 15"

 
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Let's try to find all real roots of <math>x^2+4=0</math>. If there was a real root (call it <math>r</math>), then it would satisfy <math>r^2+4=0</math>. Rearranging gives that <math>r^2=-4</math>. Therefore a real root <math>r</math> of <math>x^2+4=0</math> would satisfy <math>r^2<0</math>. However, the [[Trivial Inequality]] states that no real <math>r</math> satisfies <math>r^2<0</math>, which must mean that there are no real roots <math>r</math>; they are <math>\boxed{\mathrm{(E)}\text{ Non-existent.}}</math>
 
Let's try to find all real roots of <math>x^2+4=0</math>. If there was a real root (call it <math>r</math>), then it would satisfy <math>r^2+4=0</math>. Rearranging gives that <math>r^2=-4</math>. Therefore a real root <math>r</math> of <math>x^2+4=0</math> would satisfy <math>r^2<0</math>. However, the [[Trivial Inequality]] states that no real <math>r</math> satisfies <math>r^2<0</math>, which must mean that there are no real roots <math>r</math>; they are <math>\boxed{\mathrm{(E)}\text{ Non-existent.}}</math>
  
===Solution 3===
+
===Solution 3 (Easier Solution)===
 
Let us try to factor <math>x^2+4.</math> 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 <math>\boxed{\mathrm{(E)}\text{ Non-existent.}}</math>
 
Let us try to factor <math>x^2+4.</math> 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 <math>\boxed{\mathrm{(E)}\text{ Non-existent.}}</math>
  

Latest revision as of 07:55, 1 September 2021

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 $\boxed{\mathrm{(E)}\text{ Non-existent.}}$

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AHSME Problems and Solutions

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