Difference between revisions of "1951 AHSME Problems/Problem 47"
(Created page with "== Problem == If <math>r</math> and <math>s</math> are the roots of the equation <math>ax^2+bx+c=0</math>, the value of <math> \frac{1}{r^{2}}+\frac{1}{s^{2}} </math> is: <math>...") |
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Plugging in the values for <math>r+s</math> and <math>rs</math> gives | Plugging in the values for <math>r+s</math> and <math>rs</math> gives | ||
| − | <cmath>\frac{1}{r^2+\frac{1}{s^2}=\frac{(-b/a)^2-2(c/a)}{(c/a)^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}</cmath> | + | <cmath>\frac{1}{r^2}+\frac{1}{s^2}=\frac{(-b/a)^2-2(c/a)}{(c/a)^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}</cmath> |
== See Also == | == See Also == | ||
Revision as of 15:15, 15 May 2012
Problem
If 
and 
are the roots of the equation 
, the value of 
is:
Solution
and can be found in terms of 
, 
, and 
by using the quadratic formula; the roots are
![\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]](http://latex.artofproblemsolving.com/b/c/b/bcb2a47ec2229e2d6ee27a928e7f4d9c3c6cc06e.png)
It's not hard to check that 
and 
. Now let's algebraically manipulate what we want to find:
![\[\frac{1}{r^2}+\frac{1}{s^2}=\frac{r^2+s^2}{r^2s^2}=\frac{(r+s)^2-2rs}{(rs)^2}\]](http://latex.artofproblemsolving.com/b/0/c/b0c3e496d71654f1cfb42d83c9b7bf95bc783d9b.png)
Plugging in the values for 
and 
gives
![\[\frac{1}{r^2}+\frac{1}{s^2}=\frac{(-b/a)^2-2(c/a)}{(c/a)^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}\]](http://latex.artofproblemsolving.com/b/0/5/b05e1b7e2647ef9babb22f26d12d01c2fb29378b.png)
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 46 |
Followed by Problem 48 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
