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1955 AHSME Problems/Problem 27: Difference between revisions

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If <math>r</math> and <math>s</math> are the roots of <math>x\^{}2 - px + q = 0</math>, then <math>x\^{}2 + s\^{}2</math> equals:
== Problem 27==


<math>(A) p\^{}2 + 2q</math> <math>(B) p\^{}2 - 2q</math> <math>(C) p\^{}2 + q\^{}2</math> <math>(D) p\^{}2 - q\^{}2</math> <math>(E) p\^{}2</math>
If <math>r</math> and <math>s</math> are the roots of <math>x^2-px+q=0</math>, then <math>r^2+s^2</math> equals:
 
<math> \textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2 </math>
 
 
== Solution ==
We can write <math>r^2+s^2</math> in terms of the sum of the roots <math>(r+s)</math> and the products of the roots <math>(rs):</math>
<cmath>r^2 + s^2 = (r+s)^2 - 2rs = p^2 - 2q</cmath>
The answer is <math>\boxed{\textbf{(B)}}.</math>
== See Also ==
{{AHSME 50p box|year=1955|num-b=26|num-a=28}}
 
{{MAA Notice}}

Latest revision as of 09:47, 15 February 2021

Problem 27

If $r$ and $s$ are the roots of $x^2-px+q=0$, then $r^2+s^2$ equals:

$\textbf{(A)}\ p^2+2q\qquad\textbf{(B)}\ p^2-2q\qquad\textbf{(C)}\ p^2+q^2\qquad\textbf{(D)}\ p^2-q^2\qquad\textbf{(E)}\ p^2$


Solution

We can write $r^2+s^2$ in terms of the sum of the roots $(r+s)$ and the products of the roots $(rs):$ \[r^2 + s^2 = (r+s)^2 - 2rs = p^2 - 2q\] The answer is $\boxed{\textbf{(B)}}.$

See Also

1955 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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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