Difference between revisions of "1951 AHSME Problems/Problem 47"
m (→Solution) |
Namelyorange (talk | contribs) |
||
| (3 intermediate revisions by 2 users not shown) | |||
| Line 5: | Line 5: | ||
<math> \textbf{(E)}\ \text{none of these} </math> | <math> \textbf{(E)}\ \text{none of these} </math> | ||
| − | == Solution == | + | == Solution 1 == |
| + | Note that <math>\frac{1}{r^2}+\frac{1}{s^2} = \frac{r^2+s^2}{r^2s^2} = \frac{(r+s)^2-2(rs)}{(rs)^2}</math>. | ||
| + | |||
| + | By [[Vieta's]], this is <math>\frac{(-\frac{b}{a})^2-2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2-2ac}{c^2} \implies \boxed{(D)}</math> | ||
| + | |||
| + | == Solution 2 == | ||
<math>r</math> and <math>s</math> can be found in terms of <math>a</math>, <math>b</math>, and <math>c</math> by using the quadratic formula; the roots are | <math>r</math> and <math>s</math> can be found in terms of <math>a</math>, <math>b</math>, and <math>c</math> by using the quadratic formula; the roots are | ||
| Line 17: | Line 22: | ||
<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> | ||
| + | |||
| + | == Solution 3 == | ||
| + | here <math>r</math> and <math>s</math> are the roots of the equation <math>ax^2+bx+c=0</math>, | ||
| + | |||
| + | now we can convert this equation with roots <math>r^2</math> and <math>s^2</math>. | ||
| + | |||
| + | let, <math>y=x^2</math> then above equation becomes | ||
| + | |||
| + | <cmath>ay+c=-b\sqrt y </cmath> square on both sides we get <cmath>a^2y^2 +(2ac-b^2)y +c^2 =0 </cmath> | ||
| + | |||
| + | Again we can change this equation with roots <math>\frac{1}{r^2}</math> and <math>\frac{1}{s^2}</math> . | ||
| + | |||
| + | let <math>z=\frac{1}{y}</math> then, <math>a^2\frac{1}{z^2}+\frac{2ac-b^2}{z}+c^2=0</math> then | ||
| + | <cmath>c^2z^2+z(2ac-b^2)+a^2=0</cmath> | ||
| + | |||
| + | then the sum of roots of the above equation is <math>\frac{1}{r^2}+\frac{1}{s^2}=\frac{b^2-2ac}{c^2}</math> | ||
| + | |||
| + | hence, <cmath>\frac{1}{r^2}+\frac{1}{s^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}</cmath> | ||
== See Also == | == See Also == | ||
Latest revision as of 22:24, 16 August 2023
Problem
If 
and 
are the roots of the equation 
, the value of 
is:
Solution 1
Note that 
.
By Vieta's, this is 
Solution 2
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)
By Vieta's Formula, 
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)
Solution 3
hereand
are the roots of the equation
,
now we can convert this equation with rootsand
.
and 
.
let,then above equation becomes
then above equation becomes
square on both sides we get
![\[a^2y^2 +(2ac-b^2)y +c^2 =0\]](//latex.artofproblemsolving.com/8/e/3/8e3ceb3c6ab1d6603c8971dffa68657150d73155.png)
![\[ay+c=-b\sqrt y\]](http://latex.artofproblemsolving.com/3/c/3/3c31a91d833d10648ab0ffd60422aa4cf4605f35.png)
square on both sides we get ![\[a^2y^2 +(2ac-b^2)y +c^2 =0\]](http://latex.artofproblemsolving.com/8/e/3/8e3ceb3c6ab1d6603c8971dffa68657150d73155.png)
Again we can change this equation with rootsand
.
and 
.
let 
then, 
then
![\[c^2z^2+z(2ac-b^2)+a^2=0\]](http://latex.artofproblemsolving.com/d/4/5/d4564764a1570854c116ffea2f6d1fb54190094f.png)
then the sum of roots of the above equation is 
hence,![\[\frac{1}{r^2}+\frac{1}{s^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}\]](//latex.artofproblemsolving.com/c/c/d/ccd95ffc5f1413e967c742cf5c094c04819943e0.png)
![\[\frac{1}{r^2}+\frac{1}{s^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}\]](http://latex.artofproblemsolving.com/c/c/d/ccd95ffc5f1413e967c742cf5c094c04819943e0.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
