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

(Solution)
(Solutions to Problem 18)
Line 4: Line 4:
 
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  ==
+
==Solutions==
  
No solution available
+
===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]]

Revision as of 17:00, 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 $r$?

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 TB1 TB2 TB3 TB4
Invalid username
Login to AoPS