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Big Ideas Math Algebra 2, 2014

BI

Big Ideas Math Algebra 2, 2014 View details

4. Solving Radical Equations and Inequalities

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Exercise 5 Page 264

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

x=0 and 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(10x+9)=x+3Let's notice that in this equation on the left-hand side there is an isolated radical with index equal to 2. Then, we will raise each side of the equation to the power of 2.

sqrt(10x+9)=x+3

LHS^2=RHS^2

(sqrt(10x+9))^2=(x+3)^2

▼

Simplify

( sqrt(a) )^2 = a

10x+9=(x+3)^2

ExpandPosPerfectSquare

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

10x+9=x^2+2(x)(3)+3^2

Multiply

10x+9=x^2+6x+3^2

Calculate power

10x+9=x^2+6x+9

LHS-10x=RHS-10x

9=x^2-4x+9

LHS-9=RHS-9

0=x^2-4x

Rearrange equation

x^2-4x=0

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

x^2-4x=0

Factor out x

x(x-4)=0

Use the Zero Product Property

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

(II): LHS+4=RHS+4

lx=0 x=4

Checking the Solutions

Next, we will check the solutions by substituting 0 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 0.

sqrt(10x+9)=x+3

x= 0

sqrt(10( 0)+9)? = 0+3

▼

Simplify

Zero Property of Multiplication

sqrt(0+9)? =0+3

Add terms

sqrt(9)? =3

Calculate root

3=3 ✓

Because our substitution produced a true statement, we know that our answer, x=0, is correct. Let's check now the 4.

sqrt(10x+9)=x+3

x= 4

sqrt(10( 4)+9)? = 4+3

▼

Simplify

Multiply

sqrt(40+9)? =4+3

Add terms

sqrt(49)? =7

Calculate root

7=7 ✓

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