2005 iTest Problems/Problem 35

Problem

How many values of $x$ satisfy the equation $(x^2 - 9x + 19)^{x^2 + 16x + 60 }= 1?$

Solution 1

This equation is satisfied when:

(1) $x^{2} + 16x + 60 = 0 \Rightarrow (x + 6)(x + 10) = 0 \Rightarrow x = -6, -10$ ;

(2) $x^{2} - 9x + 19 = 1 \Rightarrow x^{2} - 9x + 18 = 0 \Rightarrow (x - 3)(x - 6) = 0 \Rightarrow x = 3, 6$ ;

(3) $x^{2} - 9x + 19 = -1 \Rightarrow x^{2} - 9x + 20 = 0 \Rightarrow (x - 4)(x - 5) = 0 \Rightarrow x = 4, 5$

of which $x = 5$ is eliminated from (3) since the exponent $5^{2} + 16(5) + 60 = 165$ is odd and $(-1)^{165} = -1$ (a contradiction). Hence, the solution set is $x = \{-10, -6, 3, 4\} \cup \{6\}$ , or $\fbox{5}$ values.

2005 iTest (Problems, Answer Key)
Preceded by: Problem 34	Followed by: Problem 36
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

Page

Toolbox

Search

2005 iTest Problems/Problem 35

Problem

Solution 1

See Also