Difference between revisions of "2007 iTest Problems/Problem 18"
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\text{(C) }108\qquad | \text{(C) }108\qquad | ||
\text{(D) It could be }0 \text{ or } 4\qquad | \text{(D) It could be }0 \text{ or } 4\qquad | ||
− | \text{(E) It could be }0 \text{ or } 108 | + | \text{(E) It could be }0 \text{ or } 108</math> |
<math>\text{(F) }18\qquad | <math>\text{(F) }18\qquad | ||
\text{(G) }-4\qquad | \text{(G) }-4\qquad | ||
\text{(H) }-108\qquad | \text{(H) }-108\qquad | ||
− | \text{(I) It could be } 0 \text{ or } -4 | + | \text{(I) It could be } 0 \text{ or } -4</math> |
<math>\text{(J) It could be } 0 \text{ or } {-108} \qquad | <math>\text{(J) It could be } 0 \text{ or } {-108} \qquad | ||
\text{(K) It could be } 4 \text{ or } {-4}\qquad | \text{(K) It could be } 4 \text{ or } {-4}\qquad | ||
− | \text{(L) There is no such value of } r\qquad | + | \text{(L) There is no such value of } r\qquad</math> |
<math>\text{(M) } 1 \qquad | <math>\text{(M) } 1 \qquad | ||
\text{(N) } {-2} \qquad | \text{(N) } {-2} \qquad | ||
\text{(O) It could be } 4 \text{ or } -4 \qquad | \text{(O) It could be } 4 \text{ or } -4 \qquad | ||
− | \text{(P) It could be } 0 \text{ or } -2 \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 | <math>\text{(Q) It could be } 2007 \text{ or a yippy dog} \qquad | ||
− | \text{(R) } 2007 | + | \text{(R) } 2007</math> |
− | == Solution == | + | ==Solutions== |
+ | |||
+ | ===Solution 1=== | ||
+ | |||
+ | By [[Vieta's Formulas]], | ||
+ | <cmath>2a+b = 6</cmath> | ||
+ | <cmath>2ab + a^2 = 9</cmath> | ||
+ | From the first equation, <math>b = -2a+6</math>. Substitute into the second equation to get | ||
+ | <cmath>-4a^2 + 12a + a^2 = 9</cmath> | ||
+ | <cmath>-3a^2 + 12a - 9 = 0</cmath> | ||
+ | <cmath>a^2 - 4a + 3 = 0</cmath> | ||
+ | Thus, <math>a = 3</math> or <math>a = 1</math>. If <math>a = 3</math>, then <math>b = 0</math>, so by Vieta's Formulas, <math>r = 0</math>. If <math>a = 1</math>, then <math>b = 4</math>, so by Vieta's Formulas, <math>r = -4</math>. The answer is <math>\boxed{\textbf{(I)}}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | The equation of the [[polynomial]] with double root <math>a</math> and single root <math>b</math> is | ||
+ | <cmath>(x-a)^2 (x-b)</cmath> | ||
+ | Expanding the polynomial results in | ||
+ | <cmath>(x^2 - 2ax + a^2)(x-b)</cmath> | ||
+ | <cmath>x^3 - (2a+b)x^2 + (2ab+a^2)x - a^2b</cmath> | ||
+ | Since <math>p = -6</math> and <math>q = 9</math>, we can write a [[system of equations]]. | ||
+ | <cmath>2a+b = 6</cmath> | ||
+ | <cmath>2ab + a^2 = 9</cmath> | ||
+ | Solving for <math>a</math> results in <math>a = 3</math> or <math>a = 1</math>. If <math>a = 3</math>, then <math>b = 0</math>, so <math>r = 0</math>. If <math>a = 1</math>, then <math>b = 4</math>, so <math>r = -4</math>. Thus, the answer is <math>\boxed{\textbf{(I)}}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 17:01, 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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