Difference between revisions of "1956 AHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | + | Dividing the equation by <math>a\quad(a\neq0)</math> gives: <math>x^2+\frac{b}{a}x+\frac{c}{a}</math>. | |
+ | Let <math>r</math> and <math>s</math> be the roots of the equation | ||
+ | <cmath>r=\frac{1}{s}</cmath> <cmath>rs=1</cmath> | ||
− | + | From [[Vieta's formula]], <math>rs=\frac{c}{a}\Rightarrow\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}</math> | |
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− | From Vieta's, <math>rs=\frac{c}{a} | ||
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== See Also== | == See Also== | ||
{{AHSME box|year=1956|num-b=6|num-a=8}} | {{AHSME box|year=1956|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:42, 15 March 2023
Problem
The roots of the equation will be reciprocal if:
Solution
Dividing the equation by gives:
.
Let
and
be the roots of the equation
From Vieta's formula,
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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