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Core Connections Integrated II, 2015 View details

2. Section 10.2

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Exercise 85 Page 578

Practice makes perfect

a To solve the given inequality, we must first find its boundary points. To do so, we will treat the inequality as if it is an equation and then solve for x. Inequality:& (4x-2)^2 ≤ 100 Equation:& (4x-2)^2 = 100 Let's solve the equation.

(4x-2)^2=100

▼

Solve for x

SqrtEqn

sqrt(LHS)=sqrt(RHS)

4x-2=± 10

AddEqn

LHS+2=RHS+2

4x=2± 10

DivEqn

.LHS /4.=.RHS /4.

x=2± 10/4

ReduceFrac

a/b=.a /2./.b /2.

x=1± 5/2

StateSol

State solutions

lcx= 1-52 & (I) x= 1+52 & (II)

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

lx=.-4 /2. x=.6 /2.

(I), (II): Calculate quotient

lx_1=-2 x_2=3

The boundary points can be found at at x=-2 and x=3. Let's mark these numbers on a number line. Examining the equation, we see that this inequality is a non-strict inequality. Therefore, the boundary points are included in the solution set and should be filled. To determine where we should shade the number line, we will test three numbers — one in each of the intervals defined by the boundary points.

a number line with an inequality marked between x=-2 and x=3

Now, we will substitute the numbers for x in the given inequality. If the inequality holds true, we should shade this region. Otherwise, we do not shade it. |c|c|r| [-0.8em] x & (4x-2)^2≤ 100 & Evaluate [0.5em] [-1em] -4 & (4(-4)-2)^2? ≤ 100 & 324 ≰ 100 * [0.5em] [-1em] & (4( )-2)^2? ≤ 100 & 4 ≤ 100 ✓ [0.5em] [-1em] 4 & (4(4)-2)^2? ≤ 100 & 196 ≰ 100 * [0.5em] The inequality is true between the boundary points. Therefore, we will shade this region.

b Let's solve the equation.

(x-1)^2=9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x-1=± 3

AddEqn

LHS+1=RHS+1

x=1± 3

StateSol

State solutions

lx_1=-2 x_2=4

c As in Part A we have an inequality, which means we will start by treating it as an equation and solve for the boundary points.

x^2+x-20=0

AddEqn

LHS+20=RHS+20

x^2+x=20

CompleteSquare

Complete the square

x^2+x+(1/2)^2=20+(1/2)^2

▼

Solve for x

SplitIntoFactors

Split into factors

x^2+2(x)(1/2)+(1/2)^2=20+(1/2)^2

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

(x+1/2)^2=20+(1/2)^2

PowQuot

(a/b)^m=a^m/b^m

(x+1/2)^2=20+1/4

NumberToFrac

a = 4* a/4

(x+1/2)^2=80/4+1/4

AddFrac

Add fractions

(x+1/2)^2=81/4

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x+1/2=± sqrt(81/4)

SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

x+1/2=± 9/2

CalcQuot

Calculate quotient

x+0.5=± 4.5

SubEqn

LHS-0.5=RHS-0.5

x=- 0.5± 4.5

StateSol

State solutions

lx_1=-5 x_2=4

The boundary points are at x=-5 and x=4. As in part A, we will mark these numbers on a number line. Notice that this inequality is strict. Therefore, the boundary points are not included in the solutions set and should not be filled. Let's mark a few test points on the numberline.

Now, we will substitute the numbers for x in the given inequality. If the inequality holds true, we should shade this region. Otherwise, we do not shade it. |c|c|r| [-0.8em] x & x^2+x-20<0 & Evaluate [0.5em] [-1em] -7 & (-7)^2+(-7)-20? < 0 & 22 ≮ 0 * [0.5em] [-1em] & ^2+ -20? < 0 & - 22 < 0 ✓ [0.5em] [-1em] -7 & 7^2+7-20? < 0 & 36 ≮ 0 * [0.5em] The inequality is true between the boundary points. Therefore, we will shade this region.