Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1961 AHSME Problems/Problem 29"

(→‎Solution 2)
(→‎Solution 2)
 
Line 30: Line 30:
 
<cmath>f(r)=f(s)=0</cmath>
 
<cmath>f(r)=f(s)=0</cmath>
 
<cmath>f(\frac{(ar+b)-b}{a})=f(\frac{(as+b)-b}{a})=0</cmath>
 
<cmath>f(\frac{(ar+b)-b}{a})=f(\frac{(as+b)-b}{a})=0</cmath>
−
Since <math>g(ar+b)=g(as+b)=0</math> we have
+
Thus <math>ar+b</math> and <math>as+b</math> are the roots to
 
<cmath>g(x)=f(\frac{x-b}{a})</cmath>
 
<cmath>g(x)=f(\frac{x-b}{a})</cmath>
 
Plugging <math>x=\frac{x-b}{a}</math> into <math>f(x)</math> gets <math>g(x)=x^2-bx+ac \boxed{B}</math>.
 
Plugging <math>x=\frac{x-b}{a}</math> into <math>f(x)</math> gets <math>g(x)=x^2-bx+ac \boxed{B}</math>.

Latest revision as of 17:04, 23 December 2019

Problem

Let the roots of $ax^2+bx+c=0$ be $r$ and $s$. The equation with roots $ar+b$ and $as+b$ is:

$\textbf{(A)}\ x^2-bx-ac=0\qquad \textbf{(B)}\ x^2-bx+ac=0 \qquad\\ \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad \textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\ \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0$

Solution

From Vieta's Formulas, $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$ in the original quadratic.

The sum of the roots in the new quadratic is \[ar + b + as + b\] \[a(r+s) + 2b\] \[b\] The product of the roots in the new quadratic is \[(ar + b)(as + b)\] \[a^2rs + arb + asb + b^2\] \[a^2rs + ab(r+s) + b^2\] \[ac - b^2 + b^2\] \[ac\] Thus, the new quadratic is $x^2-bx+ac$. The answer is $\boxed{\textbf{(B)}}$.

Solution 2

Let $f(x)=ax^2+bx+c$ and $g(x)$ be the desired function with roots $ar+b$ and $as+b$. Applying graphical transformations, we have \[f(r)=f(s)=0\] \[f(\frac{(ar+b)-b}{a})=f(\frac{(as+b)-b}{a})=0\] Thus $ar+b$ and $as+b$ are the roots to \[g(x)=f(\frac{x-b}{a})\] Plugging $x=\frac{x-b}{a}$ into $f(x)$ gets $g(x)=x^2-bx+ac \boxed{B}$.

~ Nafer

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1961_AHSME_Problems/Problem_29&oldid=113286"