Difference between revisions of "1976 AHSME Problems/Problem 6"
(Created page with "=Problem 6= If <math>c</math> is a real number and the negative of one of the solutions of <math>x^2-3x+c=0</math> is a solution of <math>x^2+3x-c=0</math>, then the solutions...") |
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=Problem 6= | =Problem 6= | ||
If <math>c</math> is a real number and the negative of one of the solutions of <math>x^2-3x+c=0</math> is a solution of <math>x^2+3x-c=0</math>, then the solutions of <math>x^2-3x+c=0</math> are | If <math>c</math> is a real number and the negative of one of the solutions of <math>x^2-3x+c=0</math> is a solution of <math>x^2+3x-c=0</math>, then the solutions of <math>x^2-3x+c=0</math> are | ||
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<math>\textbf{(A) }1,~2\qquad \textbf{(B) }-1,~-2\qquad \textbf{(C) }0,~3\qquad \textbf{(D) }0,~-3\qquad \textbf{(E) }\frac{3}{2},~\frac{3}{2}</math> | <math>\textbf{(A) }1,~2\qquad \textbf{(B) }-1,~-2\qquad \textbf{(C) }0,~3\qquad \textbf{(D) }0,~-3\qquad \textbf{(E) }\frac{3}{2},~\frac{3}{2}</math> | ||
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=Solution= | =Solution= | ||
We let the roots of the first equation be <math>r,s</math> and the roots of the second equation be <math>s, -t</math>. By Vieta's Formulas, <math>r+s=3</math> and <math>s-t=-3</math>, <math>rs=c</math> and <math>-st=c</math>. So, <math>r=t</math>. Thus, <math>t+s=3</math>, <math>s-t=3</math>, so <math>t=0</math>, and <math>s=3\Rightarrow \textbf{(C)}</math>.~MathJams | We let the roots of the first equation be <math>r,s</math> and the roots of the second equation be <math>s, -t</math>. By Vieta's Formulas, <math>r+s=3</math> and <math>s-t=-3</math>, <math>rs=c</math> and <math>-st=c</math>. So, <math>r=t</math>. Thus, <math>t+s=3</math>, <math>s-t=3</math>, so <math>t=0</math>, and <math>s=3\Rightarrow \textbf{(C)}</math>.~MathJams | ||
{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Latest revision as of 20:11, 12 July 2020
Problem 6
If 
is a real number and the negative of one of the solutions of 
is a solution of 
, then the solutions of are
Solution
We let the roots of the first equation be 
and the roots of the second equation be 
. By Vieta's Formulas, 
and 
, 
and 
. So, 
. Thus, 
, 
, so 
, and 
.~MathJams
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