MATH 2171 Homework #1 Algebra Review

Solve the inequality. (Enter your answer using interval notation.) |x + 4| ≥ 2

(-∞,-6] U [-2,∞)

Solve the inequality. (Enter your answer using interval notation.) |x − 5| < 1

(4,6)

Rewrite the expression without using the absolute-value symbol. |1 − 7x^(2)| ___________ , if −1/√7 ≤ x ≤ 1/√7 ___________ , if x < -1/√7 or x > 1/√7

1 − 7x^(2) and 7x^(2) - 1 Determine when 1 - 7x^(2) < 0 1 < 7x^(2) x^(2) > 1/7 √x^(2) > √1/7 |x| > √1/7 x < -1/√7 or x > 1/√7

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. 2x + 9 > 7

(-1,∞) 2x + 9 > 7 2x > -2 x > -1 For the number line, it would be an open circle on the point -1 and it colors to the right to infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. 1 < 3x + 10 ≤ 34

(-3,8] 1 < 3x + 10 ≤ 34 -9 < 3x ≤ 24 -3 < x ≤ 8 For the number line, it would be an open circle on the point -3 and it colors to the right to the closed circle of 8.

Solve the inequality. (Enter your answer using interval notation.) |x + 1| ≥ 9

(-∞,-10] U [8,∞)

Rewrite the expression without using the absolute-value symbol. |x − 9| if x < 9

9 - x |x − 9| if x < 9 Since x < 9, (x - 9) is negative, therefore, |x − 9| = - (x - 9) = 9 - x

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. x^(2) < 2x + 3

(-1,3) x^(2) < 2x + 3 x^(2) - 2x - 3 = 0 (x-3) (x+1) = 0 x = 3,-1 Test: (x<-1), (-1<x<3), (x>3) True:(-1<x<3) For the number line, it would be an open circle on the point -1 and it colors to the right to the open circle of 3.

Solve the inequality. (Enter your answer using interval notation.) |x| < 8

(-8,8)

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. x^(2) < 2

(-√2,√2) x^(2) < 2 x < +-√2 -√2 < x < √2 For the number line, it would be an open circle on the point ~-1.4 and it colors to the right to the open circle of ~1.4.

Solve the inequality. (Enter your answer using interval notation.) |x| ≥ 3

(-∞,-3] U [3,∞)

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. (2x + 9)(x − 4) ≥ 0

(-∞,-9/2] U [4,∞) x=(-9/2), 4 Test: (x<[-9/2]), ([-9/2]<x<4), (x>4) True: (x≤[-9/2]), (x≥4) For the number line, it would be a closed circle on the point -4.5 and it colors to the left to infinity. Then another closed circle on the point 4 colored to the right to infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. x^(2) ≥ 3

(-∞,-√3] U [√3 ,∞) x^(2) ≥ 3 x ≥ +-√3 x ≥ √3 or x ≤ -√3 For the number line, it would be a closed circle on the point ~-1.7 and it colors to the left to infinity. Then another closed circle on the point ~1.7 and it colors to the right to infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. (1/x) < 4

(-∞,0) U (1/4,∞) For the number line, it would be an open circle on the point 0 and it colors to the left to infinity. Then an open circle on the point .25 and it colors to the right to infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. x^(5) − x^(4) ≤ 0

(-∞,1] x^(5) − x^(4) = 0 x^(4) [x-1] = 0 x = 0, 1 Test: (x<0), (0<x<1), (x>1) True:(x<0), (0<x<1) Combine: x≤1 For the number line, it would be a closed circle on the point 1 and it colors to the left to infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. (x − 6)(x − 9) > 0

(-∞,6)U(9,∞) x=6,9 Test: (x<6), (6<x<9), (x>9) True: (x<6), (x>9) For the number line, it would be an open circle on the point 6 and it colors to the left to infinity. Then another open circle on point 9 and it colors to the right to infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. 3x − 25 < 2

(-∞,9) 3x - 25 < 2 3x < 27 x < 9 For the number line, it would be an open circle on point 9 and it colors to the left to negative infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. x^(2) + x + 5 > 0

(-∞,∞) x^(2) + x + 5 = 0 (-1+-√1^2 - 4 * 5) / 2 x = (-1+-i√19)/2 Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and x^(2) + x + 5 is always greater than 0. For the number line, it would be colored from negative infinity to positive infinity.

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. 1 + 9x > 9 − 3x

(2/3, ∞) 1 + 9x > 9 - 3x 9x + 3x > 9 - 1 12x > 8 x > 8/12 x > 2/3 For the number line, it would be an open circle on the point 2/3 and it colors to the right to infinity.

Rewrite the expression without using the absolute-value symbol. |4x − 3| _________ , if x ≥ 3/4 _________ , if x < 3/4

4x - 3 & 3 - 4x |4x - 3| if x ≥ 3/4 Since x ≥ 3/4, (4x - 3) is positive, therefore, |4x - 3| = 4x - 3 |4x - 3| if x < 3/4 Since x < 3/4, (4x - 3) is negative, therefore, |4x - 3| = -(4x - 3) = 3 - 4x

Solve the inequality in terms of intervals and illustrate the solution set on the real number line. (x + 2)(x − 7)(x + 4) ≥ 0

[-4,-2] U [7,∞) x = -2, 7, -4 Test: (x<-4), (-4<x<-2), (-2<x<7), (x>7) True: -4≤x≤-2 or x≥7 For the number line, it would be a closed circle on the point -4 and it colors to the right to the closed circle of -2. Then a closed circle on the point 7 and it colors to the right to infinity.

Rewrite the expression without using the absolute-value symbol. |x + 9| ________ , if x ≥ −9 ________ , if x < −9

x + 9 & -(x + 9) |x + 9| if x ≥ −9 Since x ≥ −9, (x + 9) is positive, therefore, |x + 9| = x + 9 |x + 9| if x < −9 Since x < -9, (x + 9) is negative, therefore, |x + 9| = - (x + 9)

Rewrite the expression without using the absolute-value symbol. |x − 3| if x > 3

x - 3 |x − 3| if x > 3 Since x > 3, (x - 3) is positive, therefore, |x − 3| = x - 3

Solve the equation for x. (Enter your answers as a comma-separated list.) |3x + 5| = 1 x = ________

x = -4/3 , -2

Solve the equation for x. (Enter your answers as a comma-separated list.) |2x| = 3 x = ________

x = 3/2, -3/2

Solve the equation for x. (Enter your answers as a comma-separated list.) |x + 6| = |2x + 1| x = _________

x = 5, -7/3

Rewrite the expression without using the absolute-value symbol. |x^(2) + 3|

x^(2) + 3 since x^(2) + 3 ≥ 0 for all x