1955 AHSME Problems/Problem 19

Revision as of 21:24, 5 August 2020 by Angrybird029 (talk | contribs)

Problem 19

Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:

$\textbf{(A)}\ x^2-6x+7=0\qquad\textbf{(B)}\ x^2-6x-7=0\qquad\textbf{(C)}\ x^2+6x-8=0\\ \textbf{(D)}\ x^2-6x+8=0\qquad\textbf{(E)}\ x^2+6x-7=0$

Solution

The first two hints can be expressed as the following system of equations: \[\begin{cases} (1) & a + b = 6 \\ (2) & a - b = 8 \end{cases}\] From this, we can clearly see that $a = 7$, and that $b = -1$.

Since quadratic equations can generally be expressed in the form of $(x - a)(x - b) = 0$, where a and b are roots, the correct quadratic, once factored, would look like $(x - 7)(x + 1) = 0$

Expanding the above equation gets us $\textbf{(B)} x^2 - 6x - 7 = 0$

See Also

1955 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS