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Big Ideas Math Algebra 1, 2015

BI

Big Ideas Math Algebra 1, 2015 View details

3. Solving Radical Equations

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Exercise 78 Page 566

Rearrange the radical equation so that one of the radical expressions is isolated. Then square both sides of the equation.

No solution.

Practice makes perfect

We will solve the given equation and check the solutions. Let's do these two things one at a time.

Finding the Solutions

To solve the given radical equation, we will first rearrange it so that one of the radical expressions is isolated. Then, we will square both sides of the equation.

sqrt(20-4z)+2sqrt(- z)=10

LHS-2sqrt(- z)=RHS-2sqrt(- z)

sqrt(20-4z)=10-2sqrt(- z)

LHS^2=RHS^2

( sqrt(20-4z) )^2= ( 10-2sqrt(- 4z) )^2

( sqrt(a) )^2 = a

20-4z= ( 10-2sqrt(- 4z) )^2

▼

Simplify right-hand side

ExpandNegPerfectSquare

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

20-4z= 10^2 - 40sqrt(- 4z) + ( 2sqrt(- 4z) )^2

(a b)^m=a^m b^m

20-4z= 10^2 - 40sqrt(- 4z) + 2^2 ( sqrt(- 4z) )^2

( sqrt(a) )^2 = a

20-4z= 10^2 - 40sqrt(- 4z) + 2^2 (- 4z)

Calculate power

20-4z= 100 - 40sqrt(- 4z) + 4(- 4z)

a(- b)=- a * b

20-4z= 100 - 40sqrt(- 4z) -16z

Since we still have a radical expression on one side of the equation, we will isolate the radical and then follow the same method.

20-4z= 100 - 40sqrt(- 4z) -16z

▼

Simplify

LHS+40sqrt(- 4z)=RHS+40sqrt(- 4z)

40sqrt(- 4z) +20-4z= 100 -16z

LHS-20=RHS-20

40sqrt(- 4z)-4z= 80 -16z

LHS+4z=RHS+4z

40sqrt(- 4z)= 80 -12z

.LHS /4.=.RHS /4.

40sqrt(- 4z)/4=80 -12z/4

Write as a difference of fractions

40sqrt(- 4z)/4=80/4 -12z/4

MovePartNumRight

a* b/c=a/c* b

40/4sqrt(- 4z)=80/4 -12/4z

Calculate quotient

10sqrt(- 4z)= 20 -3z

LHS^2=RHS^2

( 10 sqrt(- 4z) )^2 = (20-3z)^2

▼

Simplify

ExpandNegPerfectSquare

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

( 10sqrt(- 4z) )^2 = 20^2-120z+(3z)^2

(a b)^m=a^m b^m

10^2( sqrt(- 4z) )^2 = 20^2-120z+3^2 z^2

( sqrt(a) )^2 = a

10^2( - 4z) = 20^2-120z+3^2 z^2

Calculate power

100(- 4z) = 400-120z+9z^2

a(- b)=- a * b

- 400z = 400-120z+9z^2

LHS+400z=RHS+400z

0=400+280z+9z^2

Rearrange equation

400+280z+9z^2=0

CommutativePropAdd

Commutative Property of Addition

9z^2+280z+400=0

We now have a quadratic equation, and we need to find its roots. To do it, let's identify the values of a, b, and c. 9z^2+ 280z+ 400=0 We can see that a= 9, b= 280, and c= 400. Let's substitute these values into the Quadratic Formula.

z=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

z=- 280 ±sqrt(280^2-4( 9)( 400))/2( 9)

▼

Simplify right-hand side

Calculate power

z=280±sqrt(78 400-4(9)(400))/2(9)

Multiply

z=280±sqrt(78 400-14 400)/18

Subtract terms

z=280±sqrt(64 000)/18

SplitIntoFactors

Split into factors

z=280±sqrt(6400 * 10)/18

sqrt(a* b)=sqrt(a)*sqrt(b)

z = 280 ± sqrt(6400) * sqrt(10)/18

Calculate root

z = 280 ± 80 sqrt(10)/18

Factor out 2

z = 2 ( 140 ± 40 sqrt(10) )/18

a/b=.a /2./.b /2.

z = 140 ± 40 sqrt(10)/9

Using the Quadratic Formula, we found that the solutions of the given equation are z_1= 140 +40 sqrt(10)9 and z_2= 140 - 40 sqrt(10)9. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check z_1= 140 + 40sqrt(10)9 and z_2= 140 - 40 sqrt(10)9 one at a time.

z_1=140 + 40sqrt(10)/9

Now, let's substitute z_1= 140 + 40sqrt(10)9 into the given equation.

sqrt(20-4z)+2sqrt(- z)=10

z= 140 + 40sqrt(10)/9

sqrt(20-4 * 140 + 40sqrt(10)/9) + 2sqrt(- 140 + 40 sqrt(10)/9)? =10

▼

Simplify

Rewrite 2 as sqrt(4)

sqrt(20-4 * 140 + 40sqrt(10)/9) + sqrt(4) * sqrt(- 140 + 40 sqrt(10)/9)? =10

sqrt(a)*sqrt(b)=sqrt(a* b)

sqrt(20-4 * 140 + 40sqrt(10)/9) + sqrt(- 4 * 140 + 40 sqrt(10)/9)? =10

MoveLeftFacToNum

a*b/c= a* b/c

sqrt(20-560 + 160sqrt(10)/9) + sqrt(- 4(140 + 40 sqrt(10))/9)? =10

Distribute - 4

sqrt(20-560 + 160sqrt(10)/9) + sqrt(- 560 - 160 sqrt(10)/9)? =10

a = 9* a/9

sqrt(180/9-560 + 160sqrt(10)/9) + sqrt(- 560 - 160sqrt(10)/9) ? =10

Subtract fractions

sqrt(- 380 - 160sqrt(10)/9) + sqrt(- 560 - 160sqrt(10)/9)? =10

Use a calculator

sqrt(- 98.44) + sqrt(- 118.44) ≠ 10 *

We obtained a false statement. Therefore, z_1= 140 + 40sqrt(10)9 is not a solution to the given equation.

z_2=140 - 40 sqrt(10)/9

Now, let's substitute z_2= 140 - 40 sqrt(10)9 into the original equation.

sqrt(20-4z)+2sqrt(- z)=10

z= 140 - 40sqrt(10)/9

sqrt(20-4 * 140 - 40sqrt(10)/9) + 2sqrt(- 140 - 40 sqrt(10)/9)? =10

▼

Simplify

Rewrite 2 as sqrt(4)

sqrt(20-4 * 140 - 40sqrt(10)/9) + sqrt(4) * sqrt(- 140 - 40 sqrt(10)/9)? =10

sqrt(a)*sqrt(b)=sqrt(a* b)

sqrt(20-4 * 140 - 40sqrt(10)/9) + sqrt(- 4 * 140 - 40 sqrt(10)/9)? =10

MoveLeftFacToNum

a*b/c= a* b/c

sqrt(20-560 - 160sqrt(10)/9) + sqrt(- 4(140 - 40 sqrt(10))/9)? =10

Distribute - 4

sqrt(20-560 - 160sqrt(10)/9) + sqrt(- 560 + 160 sqrt(10)/9)? =10

a = 9* a/9

sqrt(180/9-560 - 160sqrt(10)/9) + sqrt(- 560 + 160sqrt(10)/9) ? =10

Subtract fractions

sqrt(- 380 + 160sqrt(10)/9) + sqrt(- 560 + 160sqrt(10)/9)? =10

Use a calculator

sqrt(13.996) + sqrt(- 6.004) ≠ 10 *

We also obtained a false statement. Therefore, there are no solutions to the given equation.

Subchapter links

Communicate Your Answer p.559

Monitoring Progress p.560-563

Exercises p.564-566