Difference between revisions of "1974 AHSME Problems/Problem 2"

(Created page with "==Problem== Let <math> x_1 </math> and <math> x_2 </math> be such that <math> x_1\not=x_2 </math> and <math> 3x_i^2-hx_i=b </math>, <math> i=1, 2 </math>. Then <math> x_1+x_2 </m...")
 
(added category)
Line 9: Line 9:
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=1|num-a=3}}
 
{{AHSME box|year=1974|num-b=1|num-a=3}}
 +
[[Category:Introductory Algebra Problems]]

Revision as of 09:17, 30 May 2012

Problem

Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$, $i=1, 2$. Then $x_1+x_2$ equals

$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \  } \frac{b}{3} \qquad \mathrm{(D) \  } 2b \qquad \mathrm{(E) \  }-\frac{b}{3}$

Solution

Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$. By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\frac{-h}{3}=\frac{h}{3}, \boxed{\text{B}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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