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

## 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) 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
Invalid username
Login to AoPS