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Difference between revisions of "1951 AHSME Problems/Problem 47"
 
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<math> \textbf{(E)}\ \text{none of these} </math>
 
<math> \textbf{(E)}\ \text{none of these} </math>
  
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== Solution ==
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== 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
  
 
<cmath>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</cmath>
 
<cmath>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</cmath>
  
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It's not hard to check that <math>r+s=-\frac{b}{a}</math> and <math>rs=\frac{c}{a}</math>. Now let's algebraically manipulate what we want to find:
+
By Vieta's Formula, <math>r+s=-\frac{b}{a}</math> and <math>rs=\frac{c}{a}</math>. Now let's algebraically manipulate what we want to find:
  
 
<cmath>\frac{1}{r^2}+\frac{1}{s^2}=\frac{r^2+s^2}{r^2s^2}=\frac{(r+s)^2-2rs}{(rs)^2}</cmath>
 
<cmath>\frac{1}{r^2}+\frac{1}{s^2}=\frac{r^2+s^2}{r^2s^2}=\frac{(r+s)^2-2rs}{(rs)^2}</cmath>
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<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 21:24, 16 August 2023

Problem

If $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$, the value of $\frac{1}{r^{2}}+\frac{1}{s^{2}}$ is:

$\textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}}$ $\textbf{(E)}\ \text{none of these}$

Solution 1

Note that $\frac{1}{r^2}+\frac{1}{s^2} = \frac{r^2+s^2}{r^2s^2} = \frac{(r+s)^2-2(rs)}{(rs)^2}$.

By Vieta's, this is $\frac{(-\frac{b}{a})^2-2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2-2ac}{c^2} \implies \boxed{(D)}$

Solution 2

$r$ and $s$ can be found in terms of $a$, $b$, and $c$ by using the quadratic formula; the roots are

\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

By Vieta's Formula, $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}$. Now let's algebraically manipulate what we want to find:

\[\frac{1}{r^2}+\frac{1}{s^2}=\frac{r^2+s^2}{r^2s^2}=\frac{(r+s)^2-2rs}{(rs)^2}\]

Plugging in the values for $r+s$ and $rs$ gives

\[\frac{1}{r^2}+\frac{1}{s^2}=\frac{(-b/a)^2-2(c/a)}{(c/a)^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}\]

Solution 3

here $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$,

now we can convert this equation with roots $r^2$ and $s^2$.

let, $y=x^2$ then above equation becomes

\[ay+c=-b\sqrt y\] square on both sides we get \[a^2y^2 +(2ac-b^2)y +c^2 =0\]

Again we can change this equation with roots $\frac{1}{r^2}$ and $\frac{1}{s^2}$ .

let $z=\frac{1}{y}$ then, $a^2\frac{1}{z^2}+\frac{2ac-b^2}{z}+c^2=0$ then 

       \[c^2z^2+z(2ac-b^2)+a^2=0\]

then the sum of roots of the above equation is $\frac{1}{r^2}+\frac{1}{s^2}=\frac{b^2-2ac}{c^2}$ 

hence, \[\frac{1}{r^2}+\frac{1}{s^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}\]

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

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