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Core Connections Algebra 2, 2013 View details

3. Section 8.3

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Exercise 152 Page 422

Practice makes perfect

a To solve equations with a variable expression inside a square root, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!

sqrt(x)+2 = x

SubEqn

LHS-2=RHS-2

sqrt(x) = x-2

RaiseEqn

LHS^2=RHS^2

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

PowSqrt

( sqrt(a) )^2 = a

x = (x-2)^2

ExpandNegPerfectSquare

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

x = x^2 - 2(x)(2) + 2^2

CalcPowProd

Calculate power and product

x = x^2 - 4x + 4

▼

Solve for x

SubEqn

LHS-x=RHS-x

0 = x^2 - 5x + 4

RearrangeEqn

Rearrange equation

x^2 - 5x + 4 = 0

Rewrite

Rewrite -5x as - x - 4x

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

FactorOut

Factor out x

x(x-1) - 4x + 4 = 0

FactorOut

Factor out -4

x(x-1) - 4(x-1) = 0

FactorOut

Factor out (x-1)

(x-1)(x-4) = 0

Let's now apply the Zero Product Property to find the solutions to the equation.

(x-1)(x-4)=0

ZeroProdProp

Use the Zero Product Property

lcx-1=0 & (I) x-4=0 & (II)

AddEqn

(I): LHS+1=RHS+1

lx=1 x-4=0

AddEqn

(II): LHS+4=RHS+4

lx=1 x=4

We found that the solutions to the equation are 1 and 4. Finally, we will check our solutions by substituting the value into the original equation. Let's start with x=1.

sqrt(x)+2 = x

Substitute

x= 1

sqrt(1) + 2 ? = 1

▼

Evaluate left-hand side

CalcRoot

Calculate root

1 + 2 ? = 1

AddTerms

Add terms

3≠ 1 *

Since substituting x=1 created a false statement, x=1 is not a solution to the given equation. Let's now check for x=4.

sqrt(x)+2 = x

Substitute

x= 4

sqrt(4) + 2 ? = 4

▼

Evaluate left-hand side

CalcRoot

Calculate root

2 + 2 ? = 4

AddTerms

Add terms

4 = 4 ✓

This time we created a true statement. Therefore, x=4 is indeed a solution to the given equation.

b To solve equations with a variable expression inside a square root, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Note that the square root is already isolated. Let's solve the equation!

sqrt(x)+2 = sqrt(x+6)

RaiseEqn

LHS^2=RHS^2

(sqrt(x)+2)^2 = (sqrt(x+6))^2

ExpandPosPerfectSquare

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

(sqrt(x))^2+2sqrt(x)(2)+2^2 = (sqrt(x+6))^2

PowSqrt

( sqrt(a) )^2 = a

x+2sqrt(x)(2)+2^2 = x+6

CalcPowProd

Calculate power and product