Difference between revisions of "1950 AHSME Problems/Problem 15"
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===Solution 2=== | ===Solution 2=== | ||
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> | ||
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| + | ===Solution 3=== | ||
| + | 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> | ||
==See Also== | ==See Also== | ||
Revision as of 07:54, 1 September 2021
Problem
The real roots of 
are:
Solution
Solution 1
This looks similar to a difference of squares, so we can write it as 
Neither of these factors are real.
Also, looking at the answer choices, there is no way multiplying two polynomials of degree 
will result in a polynomial of degree as well. Therefore the real factors are 
Solution 2
Let's try to find all real roots of 
. If there was a real root (call it 
), then it would satisfy 
. Rearranging gives that 
. Therefore a real root of would satisfy 
. However, the Trivial Inequality states that no real satisfies , which must mean that there are no real roots ; they are
Solution 3
Let us try to factor 
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
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
