Help with quadratic and polynomial inequality please?

Question

I need help with these 4 questions... Explainations would be helpful too.

Quadratic inequality in interval notation

X^2 - 5x + 6 > or equal to 0

2x^2 < 5x + 3

Polynomial inequality in interval notation.

X^3 - 2x^2 + x > 0

X^4 < or equal to 8x^2 - 16

Answer 1

The inequality x²  - 5x + 6 ≥ 0

• Solve the related equation x²  - 5x + 6 = 0

x² - 3x - 2x + 6 = 0

x ( x - 3) - 2(x - 3) = 0

(x - 3) (x - 2) = 0

Now, there are two ways that this product could be greater than zero (positive) - both factors are positive or both factors are negative.

Case 1: Both factors are positive

x - 3 > 0 and x - 2 > 0

x > 3 and x > 2

The only condition that makes both true is x > 3.

Case 2 : Both factors are negative.

x - 3 < 0 and x - 2 < 0

x < 3 and x < 2

The only condition that makes both true is x < 2

• Since the inequality symbol is " ≥ " We inlude it into set of solutions at x = 3 and 2.

Solution of related equation x = 3,2.

For x = 3

x² - 5x + 6 ≥ 0

(3)² - 5(3) + 6 ≥ 0

0  ≥ 0

For x = 2

x² - 5x + 6 ≥ 0

(2)² - 5(2) + 6 ≥ 0

0  ≥ 0

The above two statements are true.

The above conclude that the inequality is satisfied for all x - values in (-∞, 2] and [3, ∞).

Solution {x| x ≥ 3 and x ≤ 2}

Solution of x² - 5x + 6 ≥ 0 in interval notation (-∞, 2] U [3, ∞).

Answer 2

The inequality 2x² < 5x + 3

• Solve the related equation 2x²  - 5x - 3 = 0

2x² - 6x + x - 3 = 0

2x(x - 3) + 1(x - 3) = 0

(x - 3) (2x + 1) = 0

Now, there are two ways this product could be less than zero.One factor must be negative and one must be positive.

First situation: x - 3 < 0 and 2x + 1 > 0

x < 3 and 2x > - 1

x < 3 and x > -1/2

Second situation:

x - 3 > 0 and 2x + 1 < 0

x > 3 and 2x < -1

x > 3 and x < -1/2

There are NO values for which this situation is true.

Solution  {x | -1/2 < x < 3}

Solution of 2x² < 5x + 3 in interval notation (-1/2, 3).

Answer 3

The cubic inequlity x³ - 2x² + x  > 0

Solve the related equation x³ - 2x²  + x  = 0

x (x² - 2x + 1) = 0

x (x - 1)² = 0

Solutions are x = 0 and x = 1.

Test interval   x - value     Inequality                             Conclusion

(-∞, 0)               x =  - 1         (-1)³ - 2(-1)² + 1 > 0 ⇒ - 2 > 0              False

(0, 1)                 x = 0.5         (0.5)³ - 2(0.5)² + 1 > 0 ⇒ 0.625 > 0      True

(1, ∞)                x = 2            (2)³ - 2(2)² + 1 > 0 ⇒ 1 > 0                   True

The above conclude that the inequality is satisfied for all x - values in (0, 1) and (1, ∞).

Solution {x | x  > 1 and 0 < x  < 1}

Solution of x³ - 2x² + x > 0 in interval notation (0, 1) U (1, ∞).

Answer 4

The polynomial inequality x⁴ ≤ 8x² - 16

Solve the related equation x⁴ = 8x² - 16

x⁴ - 8x² + 16 = 0

(x² - 4)² = 0

x² - 4 = 0

x² = 4

x = ± 2

Solutions are x = 2 and x = - 2.

Now test  sample values in each interval to determine whether values in the interval satisify the inequality.

Test interval   x - value       Inequality                            Conclusion

(-∞, -2)               x =  - 3         (-3)⁴ ≤ 8(-3)² - 16 ⇒81 ≤ 56          False

(-2, 2)                  x = 0            (0)⁴ ≤ 8(0)² - 16 ⇒0 ≤ -16             False

(2, ∞)                 x = 3             (3)⁴ ≤ 8(3)² - 16 ⇒81 ≤ 56            False

Since the inequality symbol is " ≤ " We inlude it into set of solutions at x = - 2 and 2.

For x  = - 2

x⁴ ≤ 8x² - 16

(-2)⁴ ≤ 8(-2)² - 16

16  ≤ 16

For x =  2

x⁴ ≤ 8x² - 16

(2)⁴ ≤ 8(2)² - 16

16  ≤ 16

The above statements are true.

Solution of x⁴ ≤ 8x² - 16 is x = -2 and x = 2.