Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 iTest Problems/Problem 23"

m (Created page with "== Problem == Find the product of the non-real roots of the equation <cmath>(x^2-3)^2+5(x^2-3)+6=0</cmath> <math>\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } -1\quad \text...")
 
(Solution to Problem 23)
Line 2: Line 2:
  
 
Find the product of the non-real roots of the equation
 
Find the product of the non-real roots of the equation
−
<cmath>(x^2-3)^2+5(x^2-3)+6=0</cmath>
+
<cmath>(x^2-3x)^2+5(x^2-3x)+6=0</cmath>
  
 
<math>\text{(A) } 0\quad
 
<math>\text{(A) } 0\quad
Line 12: Line 12:
 
\text{(G) } -3\quad
 
\text{(G) } -3\quad
 
\text{(H) } 4\quad
 
\text{(H) } 4\quad
−
\text{(I) } -4\quad \\ </math>
+
\text{(I) } -4\quad</math>
  
 
<math>\text{(J) } 5\quad
 
<math>\text{(J) } 5\quad
Line 19: Line 19:
 
\text{(M) } -6\quad
 
\text{(M) } -6\quad
 
\text{(N) } 3+2i\quad
 
\text{(N) } 3+2i\quad
−
\text{(O) } 3-2i\quad \\ </math>
+
\text{(O) } 3-2i\quad</math>
  
 
<math>\text{(P) } \frac{-3+i\sqrt{3}}{2}\quad
 
<math>\text{(P) } \frac{-3+i\sqrt{3}}{2}\quad
Line 26: Line 26:
 
\text{(S) } 12\quad
 
\text{(S) } 12\quad
 
\text{(T) } -12\quad
 
\text{(T) } -12\quad
−
\text{(U) } 42\quad \\ </math>
+
\text{(U) } 42\quad</math>
  
 
<math>\text{(V) Ying} \quad
 
<math>\text{(V) Ying} \quad
Line 32: Line 32:
  
 
== Solution ==
 
== Solution ==
 +
Let <math>a = x^2 - 3x</math>.  Substituting that in results in
 +
<cmath>a^2 + 5a + 6</cmath>
 +
Factoring that (and substituting back) leads to
 +
<cmath>(a+2)(a+3)</cmath>
 +
<cmath>(x^2-3x+2)(x^2-3x+3)</cmath>
 +
The first part is factorable into <math>(x-2)(x-1)</math>.  The second part isn’t — and the discriminant is <math>3^2 - 4 \cdot 3 = -3</math>.  With the imaginary factors found, by [[Vieta's Formulas]], the product of the non-real roots is <math>\boxed{\textbf{(F) }3}</math>.
 +
 +
==See Also==
 +
{{iTest box|year=2007|num-b=22|num-a=24}}
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 03:23, 18 June 2018

Problem

Find the product of the non-real roots of the equation \[(x^2-3x)^2+5(x^2-3x)+6=0\]

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } -1\quad \text{(D) } 2\quad \text{(E) } -2\quad \text{(F) } 3\quad \text{(G) } -3\quad \text{(H) } 4\quad \text{(I) } -4\quad$

$\text{(J) } 5\quad \text{(K) } -5\quad \text{(L) } 6\quad \text{(M) } -6\quad \text{(N) } 3+2i\quad \text{(O) } 3-2i\quad$

$\text{(P) } \frac{-3+i\sqrt{3}}{2}\quad \text{(Q) } 8\quad \text{(R) } -8\qquad \text{(S) } 12\quad \text{(T) } -12\quad \text{(U) } 42\quad$

$\text{(V) Ying} \quad \text{(W) } 207$

Solution

Let $a = x^2 - 3x$. Substituting that in results in \[a^2 + 5a + 6\] Factoring that (and substituting back) leads to \[(a+2)(a+3)\] \[(x^2-3x+2)(x^2-3x+3)\] The first part is factorable into $(x-2)(x-1)$. The second part isn’t — and the discriminant is $3^2 - 4 \cdot 3 = -3$. With the imaginary factors found, by Vieta's Formulas, the product of the non-real roots is $\boxed{\textbf{(F) }3}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 22 		Followed by:
Problem 24
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_23&oldid=95275"