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> | ||
− | == 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
By Vieta's Formula, and
. Now let's algebraically manipulate what we want to find:
Plugging in the values for and
gives
Solution 3
hereand
are the roots of the equation
,
now we can convert this equation with rootsand
.
let,then above equation becomes
square on both sides we get
![]()
Again we can change this equation with rootsand
.
let then,
then
then the sum of roots of the above equation is
hence,![]()
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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