Difference between revisions of "1955 AHSME Problems/Problem 19"
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Let the roots of the equation be <math>x</math> and <math>y</math>. Therefore, we can set up a system of equation: | Let the roots of the equation be <math>x</math> and <math>y</math>. Therefore, we can set up a system of equation: | ||
<cmath>x+y=6</cmath><cmath>|x-y|=8</cmath>Therefore, we get <math>x=7</math> and <math>y=-1</math>. So, <math>(x-7)(x+1)=\boxed{x^2-6x-7}</math> | <cmath>x+y=6</cmath><cmath>|x-y|=8</cmath>Therefore, we get <math>x=7</math> and <math>y=-1</math>. So, <math>(x-7)(x+1)=\boxed{x^2-6x-7}</math> | ||
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== See Also == | == See Also == |
Latest revision as of 10:11, 12 July 2021
Contents
Problem 19
Two numbers whose sum is and the absolute value of whose difference is are roots of the equation:
Solution
The first two hints can be expressed as the following system of equations: From this, we can clearly see that , and that .
Since quadratic equations can generally be expressed in the form of , where a and b are roots, the correct quadratic, once factored, would look like
Expanding the above equation gets us
Solution 2
Let the roots of the equation be and . Therefore, we can set up a system of equation: Therefore, we get and . So,
- kante314 -
See Also
1955 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |
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