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Core Connections Integrated III, 2015 View details

2. Section 6.2

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Exercise 58 Page 295

Practice makes perfect

a To solve this equation algebraically, we must first eliminate the square root on the left-hand side. We can do that by squaring both sides.

x + 18 = x - 2

SqrtEqn

$LHS = RHS$

x + 18 = (x - 2)^{2}

▼

Simplify

ExpandNegPerfectSquare

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

x + 18 = x^{2} - 2 (x) (2) + 2^{2}

CalcPowProd

Calculate power and product

x + 18 = x^{2} - 4 x + 4

SubEqn

$LHS - x = RHS - x$

18 = x^{2} - 5 x + 4

SubEqn

$LHS - 18 = RHS - 18$

0 = x^{2} - 5 x - 14

Now we can proceed to solve this by using the Quadratic Formula.

0 = x^{2} - 5 x - 14

UseQuadForm

Use the Quadratic Formula: $a = 1, b = - 5, c = - 14$

x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 1 ) ( - 1 4 )}{2 ( 1 )}

▼

Evaluate right-hand side

NegNeg

$- (- a) = a$

x = \frac{5 \pm ( - 5 ) ^{2} - 4 ( 1 ) ( - 1 4 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

x = \frac{5 \pm 2 5 + 5 6}{2}

AddTerms

Add terms

x = \frac{5 \pm 8 1}{2}

CalcRoot

Calculate root

x = \frac{5 \pm 9}{2}

StateSol

State solutions

x = - 4 / 2 x = 14 / 2

CalcQuot

Calculate quotient

x_{1} = - 2 x_{2} = 7

We have two solutions, one at

x = - 2

and one at

x = 7 .

Notice that when we squared the equation, there is a risk of introducing false solutions. That's why we have to test them in the original equation.

x - 2 7 x + 18 = x - 2 - 2 + 18 = ? - 2 - 2 7 + 18 = ? 7 - 2 Evaluate 4 \neq = - 4 5 = 5 Solution \times ✓

We have one valid solution at

x = 7 .

b To solve this exercise, we will use a graphing approach. However, to do that, we must first write each side of the equation as separate functions.

∣ x + 3 ∣ = - 2 x + 9 ⇓ f (x) = ∣ x + 3 ∣ and g (x) = - 2 x + 9

To graph

g (x),

we will use its

y -

intercept and slope.

Next, we will graph

f (x) .

To do that, let's consider the general equation of an absolute value function.

ABSOLUTE VALUE FUNCTION y = a ∣ x - h ∣ + k Vertical Stretch : a Horizontal Shift : h Vertical Shift : k

If we rewrite

f (x)

to match this format exactly, we can then determine the necessary transformations that we need to perform on the parent function.

y = 1 ∣ x - (- 3) ∣ + 0 Vertical Stretch : 1 Horizontal Shift : - 3 Vertical Shift : 0

As we can see, we have to shift the parent function by

3

units in the negative horizontal direction.

The graphs intersect at $(2, 5)$ which means $x = 2$ is the solution to our equation.

c To solve this system of equations, we will use the Elimination Method.

{y + 3 = 2 x^{2} - 5 x 3 x + y = 1 (I) (II)

SysEqnSub

$(II):$ Subtract $(I)$

{y + 3 = 2 x^{2} - 5 x 3 x + y - (y + 3) = 1 - (2 x^{2} - 5 x)

▼

(II):

Simplify

Distr

$(II):$ Distribute $- 1$

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

AddSubTerms

$(II):$ Add and subtract terms

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

SubEqn

$(II):$ $LHS - 1 = RHS - 1$

{y + 3 = 2 x^{2} - 5 x 3 x - 4 = - 2 x^{2} + 5 x

AddEqn

$(II):$ $LHS + 2 x^{2} = RHS + 2 x^{2}$

{y + 3 = 2 x^{2} - 5 x 2 x^{2} + 3 x - 4 = 5 x

SubEqn

$(II):$ $LHS - 5 x = RHS - 5 x$

{y + 3 = 2 x^{2} - 5 x 2 x^{2} - 2 x - 4 = 0

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

{y + 3 = 2 x^{2} - 5 x x^{2} - x - 2 = 0

Now we can proceed to solve the equation in (II) by using the Quadratic Formula.

0 = x^{2} - x - 2

UseQuadForm

Use the Quadratic Formula: $a = 1, b = - 1, c = - 2$

x = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 2 )}{2 ( 1 )}

▼

Evaluate right-hand side

NegNeg

$- (- a) = a$

x = \frac{1 \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 2 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

x = \frac{1 \pm 1 + 8}{2}

AddTerms

Add terms

x = \frac{1 \pm 9}{2}

CalcRoot

Calculate root

x = \frac{1 \pm 3}{2}

StateSol

State solutions

x = - 2 / 2 x = 4 / 2 (I) (II)

$(I), (II):$ Calculate quotient

x_{1} = - 1 x_{2} = 2

We have two solutions, one at

x = - 1

and the other at

x = 2 .

To find the corresponding

y -

coordinates, we will substitute these solutions into either function and solve for

y .

x - 1 2 3 x + y = 1 3 (- 1) + y = 1 3 (2) + y = 1 Solve for y y = 4 y = - 5

The solutions are

(- 1, 4)

and

(2, - 5) .

d Like in Part C, we will solve this by using the Elimination Method.

{y = 2 x^{2} - 8 x + 7 y = - 2 x^{2} + 12 x - 14 (I) (II)

SysEqnSub

$(II):$ Subtract $(I)$

{y = 2 x^{2} - 8 x + 7 y - y = - 2 x^{2} + 12 x - 14 - (2 x^{2} - 8 x + 7)

▼

(II):

Simplify

Distr

$(II):$ Distribute $- 1$

{y = 2 x^{2} - 8 x + 7 y - y = - 2 x^{2} + 12 x - 14 - 2 x^{2} + 8 x - 7

AddSubTerms

$(II):$ Add and subtract terms

{y = 2 x^{2} - 8 x + 7 0 = - 4 x^{2} + 20 x - 21

MultEqn

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

{y = 2 x^{2} - 8 x + 7 0 = 4 x^{2} - 20 x + 21

Now we can proceed to solve the equation in (II) by using the Quadratic Formula.

0 = 4 x^{2} - 20 x + 21

UseQuadForm

Use the Quadratic Formula: $a = 4, b = - 20, c = 21$

x = \frac{- ( - 2 0 ) \pm ( - 2 0 ) ^{2} - 4 ( 4 ) ( 2 1 )}{2 ( 4 )}

▼

Evaluate right-hand side

NegNeg

$- (- a) = a$

x = \frac{2 0 \pm ( - 2 0 ) ^{2} - 4 ( 4 ) ( 2 1 )}{2 ( 4 )}

CalcPowProd

Calculate power and product

x = \frac{2 0 \pm 4 0 0 - 3 3 6}{8}

SubTerm

Subtract term

x = \frac{2 0 \pm 6 4}{8}

CalcRoot

Calculate root

x = \frac{2 0 \pm 8}{8}

StateSol

State solutions

x = 12 / 8 x = 28 / 8 (I) (II)

$(I), (II):$ Calculate quotient

x_{1} = 1.5 x_{2} = 3.5

We have two solutions, one at

x = 1.5

and the other at

x = 3.5 .

To find the corresponding

y -

coordinates, we will substitute these solutions into either function and evaluate.

x 1.5 3.5 2 x^{2} - 8 x + 7 2 (1.5)^{2} - 8 (1.5) + 7 2 (3.5)^{2} - 8 (3.5) + 7 y - 0.5 3.5

The solutions are

(1.5, - 0.5)

and

(3.5, 3.5) .

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

Exercise 58 Page 295

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