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Help with quadratic and polynomial inequality please?

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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
asked Sep 10, 2014 in ALGEBRA 2 by anonymous

4 Answers

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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, ∞).

answered Sep 10, 2014 by david Expert
edited Sep 10, 2014 by david
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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).

answered Sep 10, 2014 by david Expert
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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 < < 1}

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

answered Sep 10, 2014 by david Expert
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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.

answered Sep 10, 2014 by david Expert

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