Difference between revisions of "2007 iTest Problems/Problem 18"
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Suppose that <math>x^3+px^2+qx+r</math> is a cubic with a double root at <math>a</math> and another root at b, where <math>a</math> and <math>b</math> are real numbers. | Suppose that <math>x^3+px^2+qx+r</math> is a cubic with a double root at <math>a</math> and another root at b, where <math>a</math> and <math>b</math> are real numbers. | ||
If <math>p=-6</math> and <math>q=9</math>, what is <math>r</math>? | If <math>p=-6</math> and <math>q=9</math>, what is <math>r</math>? | ||
+ | |||
+ | <math>\text{(A) }0\qquad | ||
+ | \text{(B) }4\qquad | ||
+ | \text{(C) }108\qquad | ||
+ | \text{(D) It could be }0 \text{ or } 4\qquad | ||
+ | \text{(E) It could be }0 \text{ or } 108</math> | ||
+ | |||
+ | <math>\text{(F) }18\qquad | ||
+ | \text{(G) }-4\qquad | ||
+ | \text{(H) }-108\qquad | ||
+ | \text{(I) It could be } 0 \text{ or } -4</math> | ||
+ | |||
+ | <math>\text{(J) It could be } 0 \text{ or } {-108} \qquad | ||
+ | \text{(K) It could be } 4 \text{ or } {-4}\qquad | ||
+ | \text{(L) There is no such value of } r\qquad</math> | ||
+ | |||
+ | <math>\text{(M) } 1 \qquad | ||
+ | \text{(N) } {-2} \qquad | ||
+ | \text{(O) It could be } 4 \text{ or } -4 \qquad | ||
+ | \text{(P) It could be } 0 \text{ or } -2 \qquad</math> | ||
+ | |||
+ | <math>\text{(Q) It could be } 2007 \text{ or a yippy dog} \qquad | ||
+ | \text{(R) } 2007</math> | ||
==Solutions== | ==Solutions== |
Latest revision as of 16:01, 10 June 2018
Problem
Suppose that is a cubic with a double root at and another root at b, where and are real numbers. If and , what is ?
Solutions
Solution 1
By Vieta's Formulas, From the first equation, . Substitute into the second equation to get Thus, or . If , then , so by Vieta's Formulas, . If , then , so by Vieta's Formulas, . The answer is .
Solution 2
The equation of the polynomial with double root and single root is Expanding the polynomial results in Since and , we can write a system of equations. Solving for results in or . If , then , so . If , then , so . Thus, the answer is .
See Also
2007 iTest (Problems) | ||
Preceded by: Problem 17 |
Followed by: Problem 19 | |
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