Difference between revisions of "1956 AHSME Problems/Problem 7"
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Dividing both sides of the equation by <math>a\quad(a\neq0)</math> gives <math>x^2+\frac{b}{a}x+\frac{c}{a}</math>. | Dividing both sides of the equation by <math>a\quad(a\neq0)</math> gives <math>x^2+\frac{b}{a}x+\frac{c}{a}</math>. | ||
− | Letting <math>r</math> and <math>s</math> be the respective roots to this quadratic, <math>\frac{1}{s}\ | + | Letting <math>r</math> and <math>s</math> be the respective roots to this quadratic, <math>r=\frac{1}{s} \Rightarrow rs=1</math>. |
From [[Vieta's]], <math>rs=\frac{c}{a}</math>, so <math>\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}</math> | From [[Vieta's]], <math>rs=\frac{c}{a}</math>, so <math>\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}</math> |
Latest revision as of 12:46, 15 March 2023
Problem
The roots of the equation will be reciprocal if:
Solution
Dividing both sides of the equation by gives
.
Letting and
be the respective roots to this quadratic,
.
From Vieta's, , so
See Also
1956 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |
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