Exercise 140 Page 532

x^2-4x+5 = - x^2+4x-1 Let's solve this equation by using the Quadratic Formula. Before we can do that though, we must move all terms to one side of the equation.

x^2-4x+5 = - x^2+4x-1

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Simplify

AddEqn

LHS+x^2=RHS+x^2

2x^2-4x+5 =4x-1

SubEqn

LHS-4x=RHS-4x

2x^2-8x+5 =-1

AddEqn

LHS+1=RHS+1

2x^2-8x+6 =0

DivEqn

.LHS /2.=.RHS /2.

x^2-4x+3 =0

Now we can solve the equation using the Quadratic formula.

x^2-4x+3=0

UseQuadForm

Use the Quadratic Formula: a = 1, b= -4, c= 3

x=-( -4) ± sqrt(( -4)^2-4( 1)( 3))/2 * 1

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Simplify right-hand side

NegNeg

- (- a)=a

x=4± sqrt((-4)^2-4(1)( 3))/2 * 1

CalcPowProd

Calculate power and product

x=4± sqrt(16-12)/2

SubTerm

Subtract term

x=4 ± sqrt(4)/2

CalcRoot

Calculate root

x=4 ± 2/2

CalcQuot

Calculate quotient

x=2 ± 1

StateSol

State solutions

lcx=2-1 & (I) x=2+1 & (II)

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

lx_1=1 x_2=3

The system has two solutions, one at x=1 and another at x=3. This means the functions intersect twice. By substituting these values into either equation, we can calculate the corresponding y-values. y= 1^2-4( 1)+5 ⇔ y=2 y= 3^2-4( 3)+5 ⇔ y=2 The system has two points of intersection, (1,2) and (3,2).