Difference between revisions of "2007 iTest Problems/Problem 18"

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

Latest revision as of 17:01, 10 June 2018

Problem

Suppose that $x^3+px^2+qx+r$ is a cubic with a double root at $a$ and another root at b, where $a$ and $b$ are real numbers. If $p=-6$ and $q=9$, what is $r$?

$\text{(A) }0\qquad \text{(B) }4\qquad \text{(C) }108\qquad \text{(D) It could be }0 \text{ or } 4\qquad \text{(E) It could be }0 \text{ or } 108$

$\text{(F) }18\qquad \text{(G) }-4\qquad \text{(H) }-108\qquad \text{(I) It could be } 0 \text{ or } -4$

$\text{(J) It could be } 0 \text{ or } {-108} \qquad \text{(K) It could be } 4 \text{ or } {-4}\qquad \text{(L) There is no such value of } r\qquad$

$\text{(M) } 1 \qquad \text{(N) } {-2} \qquad  \text{(O)  It could be } 4 \text{ or } -4 \qquad \text{(P)  It could be } 0 \text{ or } -2 \qquad$

$\text{(Q)  It could be } 2007 \text{ or a yippy dog} \qquad \text{(R)  } 2007$

Solutions

Solution 1

By Vieta's Formulas, \[2a+b = 6\] \[2ab + a^2 = 9\] From the first equation, $b = -2a+6$. Substitute into the second equation to get \[-4a^2 + 12a + a^2 = 9\] \[-3a^2 + 12a - 9 = 0\] \[a^2 - 4a + 3 = 0\] Thus, $a = 3$ or $a = 1$. If $a = 3$, then $b = 0$, so by Vieta's Formulas, $r = 0$. If $a = 1$, then $b = 4$, so by Vieta's Formulas, $r = -4$. The answer is $\boxed{\textbf{(I)}}$.

Solution 2

The equation of the polynomial with double root $a$ and single root $b$ is \[(x-a)^2 (x-b)\] Expanding the polynomial results in \[(x^2 - 2ax + a^2)(x-b)\] \[x^3 - (2a+b)x^2 + (2ab+a^2)x - a^2b\] Since $p = -6$ and $q = 9$, we can write a system of equations. \[2a+b = 6\] \[2ab + a^2 = 9\] Solving for $a$ results in $a = 3$ or $a = 1$. If $a = 3$, then $b = 0$, so $r = 0$. If $a = 1$, then $b = 4$, so $r = -4$. Thus, the answer is $\boxed{\textbf{(I)}}$.

See Also

2007 iTest (Problems)
Preceded by:
Problem 17
Followed by:
Problem 19
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