Difference between revisions of "1950 AHSME Problems/Problem 15"
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==Solution== | ==Solution== | ||
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This looks similar to a difference of squares, so we can write it as <math>(x+2i)(x-2i).</math> Neither of these factors are real. | This looks similar to a difference of squares, so we can write it as <math>(x+2i)(x-2i).</math> Neither of these factors are real. | ||
Also, looking at the answer choices, there is no way multiplying two polynomials of degree <math>2</math> will result in a polynomial of degree <math>2</math> as well. Therefore the real factors are <math>\boxed{\mathrm{(E)}\text{ Non-existent.}}</math> | Also, looking at the answer choices, there is no way multiplying two polynomials of degree <math>2</math> will result in a polynomial of degree <math>2</math> as well. Therefore the real factors are <math>\boxed{\mathrm{(E)}\text{ Non-existent.}}</math> | ||
| − | ==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> | ||
Revision as of 14:26, 17 April 2012
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
See Also
| 1950 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
