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Difference between revisions of "1977 AHSME Problems/Problem 23"

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\textbf{(D) }\left(\frac{m}{n}\right)^2=\frac{p}{q}\qquad  
 
\textbf{(D) }\left(\frac{m}{n}\right)^2=\frac{p}{q}\qquad  
 
\textbf{(E) }\text{none of these}    </math>
 
\textbf{(E) }\text{none of these}    </math>
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== Solution ==
 
== Solution ==
  
 
Let <math>r_1</math> and <math>r_2</math> be the roots of the equation <math>x^2+mx+n=0</math>. Then, <math>r_1^3</math> and <math>r_2^3</math> are the roots of <math>x^2+px+q=0</math>. Applying [[Vieta's Formulas]] to the first equation, we have <math>r_1 + r_2 = -m</math> and <math>r_1r_2 = n</math>. Similarly, from the second equation, we have <math>r_1^3 + r_2^3 = -p</math> and <math>r_1^3r_2^3 = q</math>. Cubing <math>r_1 + r_2 = -m</math>, we get <math>r_1^3 + r_2^3 + 3r_1r_2(r_1+r_2) = -m^3</math>. Plugging in from the other equations gives <math>-p + 3n(-m)=-m^3</math>. Rearranging we have <math>\boxed{\textbf{(B)} p=m^3-3mn.}</math>
 
Let <math>r_1</math> and <math>r_2</math> be the roots of the equation <math>x^2+mx+n=0</math>. Then, <math>r_1^3</math> and <math>r_2^3</math> are the roots of <math>x^2+px+q=0</math>. Applying [[Vieta's Formulas]] to the first equation, we have <math>r_1 + r_2 = -m</math> and <math>r_1r_2 = n</math>. Similarly, from the second equation, we have <math>r_1^3 + r_2^3 = -p</math> and <math>r_1^3r_2^3 = q</math>. Cubing <math>r_1 + r_2 = -m</math>, we get <math>r_1^3 + r_2^3 + 3r_1r_2(r_1+r_2) = -m^3</math>. Plugging in from the other equations gives <math>-p + 3n(-m)=-m^3</math>. Rearranging we have <math>\boxed{\textbf{(B)} p=m^3-3mn.}</math>

Latest revision as of 22:55, 1 January 2024

If the solutions of the equation $x^2+px+q=0$ are the cubes of the solutions of the equation $x^2+mx+n=0$, then

$\textbf{(A) }p=m^3+3mn\qquad \textbf{(B) }p=m^3-3mn\qquad \textbf{(C) }p+q=m^3\qquad\\ \textbf{(D) }\left(\frac{m}{n}\right)^2=\frac{p}{q}\qquad \textbf{(E) }\text{none of these}$

Solution

Let $r_1$ and $r_2$ be the roots of the equation $x^2+mx+n=0$. Then, $r_1^3$ and $r_2^3$ are the roots of $x^2+px+q=0$. Applying Vieta's Formulas to the first equation, we have $r_1 + r_2 = -m$ and $r_1r_2 = n$. Similarly, from the second equation, we have $r_1^3 + r_2^3 = -p$ and $r_1^3r_2^3 = q$. Cubing $r_1 + r_2 = -m$, we get $r_1^3 + r_2^3 + 3r_1r_2(r_1+r_2) = -m^3$. Plugging in from the other equations gives $-p + 3n(-m)=-m^3$. Rearranging we have $\boxed{\textbf{(B)} p=m^3-3mn.}$

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