Difference between revisions of "1955 AHSME Problems/Problem 19"
Angrybird029 (talk | contribs) (Created page with "== Problem 19== Two numbers whose sum is <math>6</math> and the absolute value of whose difference is <math>8</math> are roots of the equation: <math> \textbf{(A)}\ x^2-6x+...") |
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Expanding the above equation gets us <math>\textbf{(B)} x^2 - 6x - 7 = 0</math> | Expanding the above equation gets us <math>\textbf{(B)} x^2 - 6x - 7 = 0</math> | ||
| + | == See Also == | ||
| + | {{AHSME box|year=1955|num-b=18|num-a=20}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 21:24, 5 August 2020
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:
![\[\begin{cases} (1) & a + b = 6 \\ (2) & a - b = 8 \end{cases}\]](http://latex.artofproblemsolving.com/6/7/1/671fe691566d8f06fafc1222d47583957f1ce6ea.png)
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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
