Intermediate Algebra Review Notes Exam #4

Solutions to Exercises

Exercise 1(a)

Exercise 1(b)

Exercise 2(a)

Exercise 2(b)

Exercise 3(a)

Write



in the equivalent form

2x - 5 < -11 or 2x - 5 > 11

Solve each side independently. Add 5, then divide by 2.

In interval notation, the solution is (-∞,-3) ∪ (8,∞).

Exercise 3(b)

Multipy each side

by 3.

Write this last result in the equivalent form

-3 < 2x - 5 < 3. First add 5 to all three sides, then divide all three sides by 2.

Solve.

2 < 2x < 8
1 < x < 4

In interval notation, the solution is (1, 4).

Exercise 4(a) Isolate the absolute values on one side of the equation.



Divide by -3, reversing the inequality.

Write in equivalent form and solve.

Exercise 4(b) Isolate the absolute value.



Divide both sides by -4, reversing the inequality.



Write as an equivalent inequality and solve .

or

Exercise 5(a) To place f(x) = x² -6x-16 in vertex form , take one-half of the middle coefficient
and square , i.e., [(1/2)(-6)]² = 9. Add and subtract this amount , factor and simplify .

f(x) = x² - 6x + 9 - 9 - 16
f(x) = (x - 3)² - 25

The parabola opens upward, the vertex is at (3,-25), and the equation of the axis of symmetry is
x = 3.

Because f(0) = -16, the y-intercept is (0,-16). To find the x-intercept, set y = 0.

0 = x² - 6x - 16

Note that ac = (1)(-16) = -16. The integer pair {2,-8} has product -16 and sums to -6. Hence,

0 = (x + 2)(x - 8).

Set each factor equal to zero and solve .
 

Thus, the x-intercepts are (-2, 0) and (8, 0).

Exercise 5(b) To place f(x) = -2x² + 13x + 24 in vertex form, first factor out a -2.

Next, take one-half of the middle coefficient and square , i.e., [(1/2)(-13/2)] = 169/16. Add and
subtract this amount, factor and simplify .

The parabola opens downward, the vertex is at (13/4, 361/8), and the equation of the axis of
symmetry is x = 13/4.

Because f(0) = 24, the y-intercept is (0, 24). To find the x-intercept, set y = 0, then multiply
both sides by -1.

0 = -2x² + 13x + 24
0 = 2x² - 13x - 24

Note that ac = (2)(-24) = -48. The integer pair {3,-16} has product -48 and sums to -13.
Hence,

0 = 2x² + 3x - 16x - 24
0 = x(2x + 3) - 8(2x + 3)
0 = (x - 8)(2x + 3).

Set each factor equal to zero and solve.

Thus, the x-intercepts are (-3/2, 0) and (8, 0).