Difference between revisions of "1954 AHSME Problems/Problem 41"

(Created page with "== Problem 41== The sum of all the roots of <math>4x^3-8x^2-63x-9=0</math> is: <math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 ...")
 
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<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0  </math>   
 
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0  </math>   
 
    
 
    
== Solution ==
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== Solution 1 ==
 
By Vieta's Formulas, <math>\frac{--8}{4}=2</math>, <math>\fbox{D}</math>
 
By Vieta's Formulas, <math>\frac{--8}{4}=2</math>, <math>\fbox{D}</math>
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== Solution 2 ==
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<math>4x^3-8x^2-63x-9=0</math>
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<math>4(x^3-2x^2-\frac{63}{4}x-\frac{9}{4})=0</math>
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By Vieta's Formulas:
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<math>x^3+ax^2+bx+c=(x-r)(x-p)(x-q)</math>
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<math>x^3+ax^2+bx+c=x^3+(-r-p-q)x^2+(rp+pq+rq)x+(-rpq)</math>
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<math>x^3-2x^2-\frac{63}{4}x-\frac{9}{4}=x^3+(-r-p-q)x^2+(rp+pq+rq)x+(-rpq)</math>
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We can see that the negative sum of the roots is the coefficient of the <math>x^2</math> term, <math>-2</math>.
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So the actual sum of the roots is <math>-(-2)</math>, or <math>2 \implies \fbox{D}</math>

Revision as of 16:10, 15 April 2017

Problem 41

The sum of all the roots of $4x^3-8x^2-63x-9=0$ is:

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0$

Solution 1

By Vieta's Formulas, $\frac{--8}{4}=2$, $\fbox{D}$

Solution 2

$4x^3-8x^2-63x-9=0$

$4(x^3-2x^2-\frac{63}{4}x-\frac{9}{4})=0$

By Vieta's Formulas: $x^3+ax^2+bx+c=(x-r)(x-p)(x-q)$

$x^3+ax^2+bx+c=x^3+(-r-p-q)x^2+(rp+pq+rq)x+(-rpq)$

$x^3-2x^2-\frac{63}{4}x-\frac{9}{4}=x^3+(-r-p-q)x^2+(rp+pq+rq)x+(-rpq)$

We can see that the negative sum of the roots is the coefficient of the $x^2$ term, $-2$. So the actual sum of the roots is $-(-2)$, or $2 \implies \fbox{D}$

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