Exercise 90 Page 607

We want to solve the given system of equations using the Substitution Method. x^2+y^2=16 & (I) y=x^2-4 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value x^2-4 for y in Equation (I).

x^2+y^2=16 y=x^2-4

Substitute

(I): y= x^2-4

x^2+( x^2-4)^2=16 y=x^2-4

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(I): Simplify

ExpandPosPerfectSquare

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

x^2+(x^2)^2-2(x^2)(4)+4^2=16 y=x^2-4

CalcPowProd

(I): Calculate power and product

x^2+x^4-8x^2+16=16 y=x^2-4

SubEqn

(I): LHS-16=RHS-16

x^2+x^4-8x^2=0 y=x^2-4

SubTerm

(I): Subtract term

x^4-7x^2=0 y=x^2-4

Notice that in Equation (I) we have a fourth order polynomial equation in terms of only the x-variable. Since there are no constant terms, it seems that trying to solve it by factoring will be a good idea. Let's do it!

x^4-7x^2=0

FactorOut

Factor out x^2

x^2(x^2-7)=0

WritePow

Write as a power

x^2(x^2-(sqrt(7))^2)=0

FacDiffSquares

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

x^2(x+sqrt(7))(x-sqrt(7))=0

Now that the equation is written in factored form, we will use the Zero Product Property to solve it.

x^2(x+sqrt(7))(x-sqrt(7))=0

ZeroProdProp

Use the Zero Product Property

lcx^2=0 & (I) x+sqrt(7)=0 & (II) x-sqrt(7)=0 & (III)

SqrtEqn

(I): sqrt(LHS)=sqrt(RHS)

lx=0 x+sqrt(7)=0 x-sqrt(7)=0

SubEqn

(II): LHS-sqrt(7)=RHS-sqrt(7)

lx=0 x=- sqrt(7) x-sqrt(7)=0

AddEqn

(III): LHS+sqrt(7)=RHS+sqrt(7)

lx=0 x=- sqrt(7) x=sqrt(7)

We found that we have three solution to Equation (I): x=0, x=-sqrt(7), and x=sqrt(7). We will substitute each of these values into Equation (II) to find the respective y-values of solutions to the original system. Let's start with x=0.

y=x^2-4

Substitute

x= 0

y= 0^2-4

CalcPow

Calculate power

y=0-4

SubTerm

Subtract term

y=-4

We found that one of the solutions to our system is point (0,-4). Now, let's substitute x=-sqrt(7) into Equation (II).

y=x^2-4

Substitute

x= -sqrt(7)

y=( -sqrt(7))^2-4

NegBaseToPosBase

(- a)^2=a^2

y=(sqrt(7))^2-4

SqrtPowToNumber

sqrt(a^2)=a

y=7-4

SubTerm

Subtract term

y=3

Therefore, the second solution is the point (-sqrt(7),3). Finally, let's substitute x = sqrt(7) into Equation (II).

y=x^2-4

Substitute

x= sqrt(7)v

y=( sqrt(7))^2-4

SqrtPowToNumber

sqrt(a^2)=a

y=7-4

SubTerm

Subtract term

y=3

The last solution to our system is the point (sqrt(7),3).

Checking Our Answer

Checking the answer

We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (0,-4). We will substitute 0 and - 4 for x and y, respectively, in Equation (I) and Equation (II).

x^2+y^2=16 y=x^2-4

(I), (II): x= 0, y= - 4

0^2+( -4)^2? =16 -4? = 0^2-4

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Simplify

(I), (II): Calculate power

0+16? =16 -4? =0-4

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

16=16 ✓ - 4=- 4 ✓

Since both equations produced true statements, the solution (0,-4) is correct. Let's now check ( - sqrt(7), 3).

x^2+y^2=16 y=x^2-4

(I), (II): x= - sqrt(7), y= 3

( - sqrt(7))^2+ 3^2? =16 3? =( - sqrt(7))^2-4

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Simplify

(I), (II): Calculate power

7+9? =16 3? =7-4

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

16=16 ✓ - 4=- 4 ✓

Since again both equations produce true statements, the solution ( - sqrt(7), 3) is also correct. Finally, let's check ( sqrt(7), 3).

x^2+y^2=16 y=x^2-4

(I), (II): x= sqrt(7), y= 3

( sqrt(7))^2+ 3^2? =16 3? =( sqrt(7))^2-4

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Simplify

(I), (II): Calculate power

7+9? =16 3? =7-4

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

16=16 ✓ - 4=- 4 ✓

Once again both equations produce true statements. Therefore, ( sqrt(7), 3) is also correct.

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Checking Our Answer