Difference between revisions of "1954 AHSME Problems/Problem 41"
Katzrockso (talk | contribs) (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 == | + | == 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> | ||
| + | == Solution 2 == | ||
| + | <math>4x^3-8x^2-63x-9=0</math> | ||
| + | |||
| + | <math>4(x^3-2x^2-\frac{63}{4}x-\frac{9}{4})=0</math> | ||
| + | |||
| + | By Vieta's Formulas: | ||
| + | <math>x^3+ax^2+bx+c=(x-r)(x-p)(x-q)</math> | ||
| + | |||
| + | <math>x^3+ax^2+bx+c=x^3+(-r-p-q)x^2+(rp+pq+rq)x+(-rpq)</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | We can see that the negative sum of the roots is the coefficient of the <math>x^2</math> term, <math>-2</math>. | ||
| + | So the actual sum of the roots is <math>-(-2)</math>, or <math>2 \implies \fbox{D}</math> | ||
Revision as of 15:10, 15 April 2017
Problem 41
The sum of all the roots of 
is:
Solution 1
By Vieta's Formulas, 
, 
Solution 2
By Vieta's Formulas:
We can see that the negative sum of the roots is the coefficient of the 
term, 
.
So the actual sum of the roots is 
, or 
