College Algebra Exam 1 - Page 1 Solutions

1. Simplify to an answer containing no negative exponents: (4xy^-2)²/(2x^-1y)³ .

Solution:

(4xy-2)2 (2x-1y)3 = 42 x2 (y-2)2 23 (x-1)3y3 = 16 x2 y-4 8 x-3y3 = 2 x(2-(-3)) y-4-3 = 2 x5 y-7 = 2 x5 y7

2. Extract squares: Ö(16 (x+3)²). Check your answer at x=-4.

Solution:

_______ Ö16 (x+3)2   =   __ Ö16     _____ Ö(x+3)2   = 4 | x+3 |       (absolute value needed since Ö   is the positive square root).

Check: At x=-4

_______ Ö16 (x+3)2   =   ________ Ö16 (-4+3)2   =   ______ Ö16 (-1)2   =   __ Ö16   =4

4|x+3|=4|-4+3|=4|-1|=4.

3. Solve: x²=4x⁴.

Solution: Equation is non-linear so set one side equal to zero and factor:

4x4 = x2 (subtract x2)                 4x4-x2 = 0 (factor out common x2)             x2(4x2 -1) = 0                                                                                (second factor is difference of two squares)         x2( (2x)2 -12) = 0 x2( 2x-1)(2x+1) = 0

So solutions correspond to x=0, (2x-1)=0 or (2x+1)=0, giving

x=0,     x= 1 2 ,   or x=- 1 2 .

4. Find the domain of Ö(x+2) - 1/(x-3).

Solution: 1/(x-3) requires x ¹ 3 (no dividing through by 0, so (x-3) ¹ 0).

Ö(x+2) requires x ³ -2 (can't take square-roots of negatives, so (x+2) ³ 0).

Since both restrictions must be satisfied domain is {x| x ³ -2 and x ¹ 3}

(which could alternatively be written in the form -2 £ x < 3 or x > 3).