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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 25 Page 266

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

x=4

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(3x-8)+1=sqrt(x+5) Notice that in this equation there is an isolated radical with index equal to 2 on the right-hand side. Then, we will raise each side of the equation to the power of 2.

sqrt(3x-8)+1=sqrt(x+5)

LHS^2=RHS^2

(sqrt(3x-8)+1)^2=(sqrt(x+5))^2

▼

Simplify

( sqrt(a) )^2 = a

(sqrt(3x-8)+1)^2=x+5

ExpandPosPerfectSquare

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

(sqrt(3x-8))^2+2(sqrt(3x-8))(1)+1^2=x+5

( sqrt(a) )^2 = a

3x-8+2(sqrt(3x-8))(1)+1^2=x+5

Multiply

3x-8+2sqrt(3x-8)+1^2=x+5

Calculate power

3x-8+2sqrt(3x-8)+1=x+5

Add terms

3x-7+2sqrt(3x-8)=x+5

We obtained another radical equation. Let's now isolate the radical, sqrt(3x-8), on one side of the equation.

3x-7+2sqrt(3x-8)=x+5

▼

Isolate sqrt(3x-8)

LHS-3x=RHS-3x

- 7+2sqrt(3x-8)=- 2x+5

LHS+7=RHS+7

2sqrt(3x-8)=- 2x+12

.LHS /2.=.RHS /2.

sqrt(3x-8)=- 2x+12/2

Factor out 2

sqrt(3x-8)=2(- x+6)/2

Calculate quotient

sqrt(3x-8)=- x+6

Now, just like before, we will raise each side of the equation to the power of 2.

sqrt(3x-8)=- x+6

LHS^2=RHS^2

(sqrt(3x-8))^2=(- x+6)^2

▼

Simplify

( sqrt(a) )^2 = a

3x-8=(- x+6)^2

ExpandPosPerfectSquare

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

3x-8=(- x)^2+2(- x)(6)+6^2

Calculate power and product

3x-8=x^2-12x+36

LHS-3x=RHS-3x

- 8=x^2-15x+36

LHS+8=RHS+8

0=x^2-15x+44

Rearrange equation

x^2-15x+44=0

We got a quadratic equation. Notice that there are many ways to solve a quadratic equation. We will use the Quadratic Formula to solve the equation. Let's determine a, b, and c. x^2-15x+44=0 ⇔ 1x^2+( - 15)x+ 44=0 We see above that a= 1, b= - 15, and c= 44. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=-( - 15)±sqrt(( - 15)^2-4( 1)( 44))/2( 1)

▼

Simplify right-hand side

Calculate power

x=-(- 15)±sqrt(225-4(1)(44))/2(1)

Multiply

x=15±sqrt(225-176)/2

Subtract terms

x=15±sqrt(49)/2

Calculate root

x=15± 7/2

Now we can calculate the first root using the positive sign and the second root using the negative sign.

x=15± 7/2
x=15+7/2	x=15-7/2
x=11	x=4

Checking the Solutions

Next, we will check the solutions by substituting 11 and 4 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 11.

sqrt(3x-8)+1=sqrt(x+5)

x= 11

sqrt(3( 11)-8)+1? =sqrt(11+5)

▼

Simplify

Multiply

sqrt(33-8)+1? =sqrt(11+5)

Add and subtract terms

sqrt(25)+1? =sqrt(16)

Calculate root

5+1? =4

Add terms

6≠ 4 *

Because our substitution produced a false statement, we know that our answer, x=11, is not correct. Let's check now the 4.

sqrt(3x-8)+1=sqrt(x+5)

x= 4

sqrt(3( 4)-8)+1? =sqrt(4+5)

▼

Simplify

Multiply

sqrt(12-8)+1? =sqrt(4+5)

Add and subtract terms

sqrt(4)+1? =sqrt(9)

Calculate root

2+1? =3

Add terms

3=3 ✓

Because our substitution produced a true statement, we know that our answer, x=4, is correct.

Subchapter links

Communicate Your Answer p.261

Monitoring Progress p.262-265

Exercises p.266-268