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Big Ideas Math Algebra 2, 2014

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Big Ideas Math Algebra 2, 2014 View details

4. Solving Radical Equations and Inequalities

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Exercise 19 Page 266

Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.

x=0 and x= 12

Practice makes perfect

Solving a radical equation usually involves three main steps.

Isolate the radical on one side of the equation.
Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.
Solve the resulting equation. Remember to check your results!

Now we can analyze the given radical equation. sqrt(8x^3-1)=2x-1Notice that in this equation there is an isolated radical with index equal to 3 on the left-hand side. Then, we will raise each side of the equation to the power of 3.

sqrt(8x^3-1)=2x-1

LHS^3=RHS^3

(sqrt(8x^3-1))^3=(2x-1)^3

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Simplify

RootPowToNumber

sqrt(a^n)=a

8x^3-1=(2x-1)^3

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

8x^3-1=(2x)^3-3(2x)^2(1)+3(2x)(1)^2-(1)^3

Calculate power

8x^3-1=8x^3-3(4x^2)(1)+3(2x)(1)-1

Multiply

8x^3-1=8x^3-12x^2+6x-1

LHS+1=RHS+1

8x^3=8x^3-12x^2+6x

LHS-8x^3=RHS-8x^3

0=- 12x^2+6x

Rearrange equation

- 12x^2+6x=0

We obtained a quadratic equation. There are many ways to solve a quadratic equation. We will solve it by factoring it.

- 12x^2+6x=0

Factor out 6x

6x(- 2x+1)=0

Use the Zero Product Property

lc6x=0 & (I) - 2x+1=0 & (II)

▼

Solve for x

(I): .LHS /6.=.RHS /6.

lcx=0 & (I) - 2x+1=0 & (II)

(II): LHS-1=RHS-1

lcx=0 & (I) - 2x=- 1 & (II)

(II): .LHS /(- 2).=.RHS /(- 2).

lcx=0 & (I) x= 12 & (II)

Checking the Solutions

Next, we will check the solutions by substituting 0 and 12 for x into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution. Let's first check the 0.

sqrt(8x^3-1)=2x-1

x= 0

sqrt(8( 0)^3-1)? =2( 0)-1

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Simplify

Calculate power

sqrt(8(0)-1)? =2(0)-1

Zero Property of Multiplication

sqrt(0-1)? =0-1

Subtract terms

sqrt(- 1)? =- 1

Calculate root

- 1=- 1 ✓

Because our substitution produced a true statement, we know that our answer, x=0, is correct. Let's check now the 12.

sqrt(8x^3-1)=2x-1

x= 1/2

sqrt(8( 1/2)^3-1)? =2( 1/2)-1

▼

Simplify

Calculate power

sqrt(8(1/8)-1)? =2(1/2)-1

MoveLeftFacToNum

a*b/c= a* b/c

sqrt(8/8-1)? =2/2-1

Calculate quotient

sqrt(1-1)? =1-1

Subtract terms

sqrt(0)? =0

Calculate root

0=0 ✓

Because our substitution produced a true statement, we know that our answer, x= 12, is correct.

Subchapter links

Communicate Your Answer p.261

Monitoring Progress p.262-265

Exercises p.266-268