1961 AHSME Problems/Problem 29: Difference between revisions

Revision as of 16:54, 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

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
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@@ Line 24: / Line 24: @@
 <cmath>ac</cmath>
 Thus, the new quadratic is <math>x^2-bx+ac</math>.  The answer is <math>\boxed{\textbf{(B)}}</math>.
+== Solution 2 ==
 ==See Also==

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1961 AHSME Problems/Problem 29: Difference between revisions

Revision as of 16:54, 23 December 2019

Contents

Problem

Solution

Solution 2

See Also