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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 16 Page 266

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

x=25

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. x-10=sqrt(9x)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.

x-10=sqrt(9x)

LHS^2=RHS^2

(x-10)^2=(sqrt(9x))^2

▼

Simplify

( sqrt(a) )^2 = a

(x-10)^2=9x

ExpandNegPerfectSquare

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

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

Multiply

x^2-20x+10^2=9x

Calculate power

x^2-20x+100=9x

LHS-9x=RHS-9x

x^2-29x+100=0

We obtained a quadratic equation. There are many ways to solve a quadratic equation. We will use the Quadratic Formula to solve this equation. Let's determine a, b, and c. x^2-29x+100=0 ⇔ 1x^2+( -29)x+ 100=0 We see above that a= 1, b= - 29, and c= 100. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=-( - 29)±sqrt(( - 29)^2-4( 1)( 100))/2( 1)

▼

Simplify right-hand side

- (- a)=a

x=29±sqrt((- 29)^2-4(1)(100))/2(1)

Calculate power

x=-(- 29)±sqrt(841-4(1)(100))/2(1)

Multiply

x=29±sqrt(841-400)/2

Subtract terms

x=29±sqrt(441)/2

Calculate root

x=29± 21/2

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

x=29± 21/2
x=29+21/2	x=29-21/2
x=25	x=4

Checking the Solutions

Next, we will check the solutions by substituting 25 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 25.

x-10=sqrt(9x)

x= 25

25-10? =sqrt(9( 25))

▼

Simplify

Multiply

25-10? =sqrt(225)

Calculate root

25-10? =15

Subtract terms

15=15 ✓

Because our substitution produced a true statement, we know that our answer, x=25, is correct. Let's check now the 4.

x-10=sqrt(9x)

x= 4

4-10? =sqrt(9( 4))

▼

Simplify

Multiply

4-10? =sqrt(36)

Calculate root

4-10? =6

Subtract terms

- 6≠ 6 *

Because our substitution produced a false statement, we know that our answer, x=4, is not correct.

Subchapter links

Communicate Your Answer p.261

Monitoring Progress p.262-265

Exercises p.266-268