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Core Connections Integrated III, 2015 View details

2. Section 6.2

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Exercise 50 Page 290

a To solve this inequality, we will first consider its related equation.

I n e q u a l i t y : E q u a t i o n : x^{2} - 2 x < 3 x^{2} - 2 x = 3

By solving the related equation, we can identify the inequality's boundary point(s). Notice that these boundary point(s) will be excluded from the solution set since the inequality is strict.

x^{2} - 2 x = 3

AddEqn

$LHS + (\frac{- 2}{2})^{2} = RHS + (\frac{- 2}{2})^{2}$

x^{2} - 2 x + (\frac{- 2}{2})^{2} = 3 + (\frac{- 2}{2})^{2}

▼

Solve for

x

CalcQuot

Calculate quotient

x^{2} - 2 x + (- 1)^{2} = 3 + (- 1)^{2}

NegBaseToPosBase

$(- a)^{2} = a^{2}$

x^{2} - 2 x + 1^{2} = 3 + 1^{2}

SplitIntoFactors

Split into factors

x^{2} - 2 (x) (1) + 1^{2} = 3 + 1^{2}

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

(x - 1)^{2} = 3 + 1^{2}

BaseOne

$1^{a} = 1$

(x - 1)^{2} = 3 + 1

AddTerms

Add terms

(x - 1)^{2} = 4

SqrtEqn

$LHS = RHS$

x - 1 = \pm 2

AddEqn

$LHS + 1 = RHS + 1$

x = 1 \pm 2

StateSol

State solutions

{x = 1 - 2 x = 1 + 2 (I) (II)

$(I), (II):$ Add and subtract terms

{x = - 1 x = 3

We have two boundary points. Let's draw these on a number line. Since the original inequality is strict, they are not a part of the solution set. We show this by making them open. We will also include three test points which we will use to figure out which interval(s) to shade.

By substituting the test point's numerical value in the original inequality, we can determine where it holds true.

x - 2 04 x^{2} - 2 x < 3 (- 2)^{2} - 2 (- 2) < ? 3 (0)^{2} - 2 (0) < ? 3 (4)^{2} - 2 (4) < ? 3 Evaluate 8 ≮ 3 \times 0 < 3 ✓ 8 ≮ 3 \times

We have to shade the interval that is between the boundary points. We get the solution set

- 1 < x < 3 .

b Like in Part A, to solve this inequality, we will begin by considering the related equation instead.

I n e q u a l i t y : E q u a t i o n : 3 x - x^{2} \leq 2 3 x - x^{2} = 2

By solving the equation, we can identify the inequality's boundary point(s). Notice that theses boundary point(s) are included in the solution set since the inequality is nonstrict.

3 x - x^{2} = 2

▼

Simplify

CommutativePropAdd

Commutative Property of Addition

- x^{2} + 3 x = 2

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

x^{2} - 3 x = - 2

AddEqn

$LHS + (\frac{- 3}{2})^{2} = RHS + (\frac{- 3}{2})^{2}$

x^{2} - 3 x + (\frac{- 3}{2})^{2} = - 2 + (\frac{- 3}{2})^{2}

▼

Solve for

x

CalcQuot

Calculate quotient

x^{2} - 3 x + (- 1.5)^{2} = - 2 + (- 1.5)^{2}

NegBaseToPosBase

$(- a)^{2} = a^{2}$

x^{2} - 3 x + 1 . 5^{2} = - 2 + 1 . 5^{2}

SplitIntoFactors

Split into factors

x^{2} - 2 (x) (1.5) + 1 . 5^{2} = - 2 + 1 . 5^{2}

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

(x - 1.5)^{2} = - 2 + 1 . 5^{2}

CalcPow

Calculate power

(x - 1.5)^{2} = - 2 + 2.25

AddTerms

Add terms

(x - 1.5)^{2} = 0.25

SqrtEqn

$LHS = RHS$

x - 1.5 = \pm 0.5

AddEqn

$LHS + 1.5 = RHS + 1.5$

x = 1.5 \pm 0.5

StateSol

State solutions

{x = 1.5 - 0.5 x = 1.5 + 0.5 (I) (II)

$(I), (II):$ Add and subtract terms

{x = 1 x = 2

We have two boundary points. Let's draw these on a number line. Since the original inequality is nonstrict, they are a part of the solution set. We show this by keeping them solid. We will also include three test points which we will use to figure out which interval(s) to shade.

By substituting the test point's numerical value in the original inequality, we can determine where it holds true.

x 0.5 1.5 2.5 3 x - x^{2} \leq 2 3 (0.5) - (0.5)^{2} \leq ? 2 3 (1.5) - (1.5)^{2} \leq ? 2 3 (2.5) - (2.5)^{2} \leq ? 2 Evaluate 1.25 \leq 2 ✓ 2.25 ≰ 2 \times 1.25 \leq 2 ✓

As we can see, we have to shade the interval to the left of

x = 1

and to the right of

x = 3 .

We get the solution set

x \leq 1

x \geq 2 .

Subchapter links

6.2.1 Statistical Test Using Sampling Variability p.283-285

6.2.2 Variability in Experimental Results p.289-290

6.2.3 Quality Control p.293-295

6.2.4 Statistical Process Control p.299-301