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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...")
 
(Undo revision 99013 by Skrublord420 (talk))
 
(2 intermediate revisions by 2 users not shown)
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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
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\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
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\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
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== 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>
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<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]]

Latest revision as of 15:48, 24 November 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
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