Download ppt - § 7.6

§ 7.6

Radical Equations

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 7.6

Radical Equations

A radical equation is an equation in which the variable occurs in a square root, cube root, or any higher root. In this section, you will learn how to solve radical equations.

When the variable occurs in a square root, it is necessary to square both sides of the equation. When you square both sides of an equation, sometimes extra answers creep in, called extraneous roots. For example, consider the following very simple original equation.

2

4

22

x

x

xSquare both sides.Solve the equation. But -2 does notwork in the original equation. Then -2 is an extraneous root. It’s like a hitchhiker that we picked up when we squared both sides. It works only in the squared form and is not a root of the original.


Radical Equations

Just to note then….

When you solve a rational equation and must raise both sides to an even power, remember to check your roots. Throw out any extra (extraneous) roots from your solution set.

Another thing that you should know is … if your equation contains two or more square root expressions, you will need to isolate one square root expression, square both sides, and then you may have to repeat the process. You may have to square both sides twice.


Solving Radical Equations

Solving Radical Equations Containing nth Roots

1) If necessary, arrange terms so that one radical is isolated on one side of the equation.

2) Raise both sides of the equation to the nth power to eliminate the nth root.

3) Solve the resulting equation. If this equation still contains radicals, repeat steps 1 and 2.

4) Check all proposed solutions in the original equation.



Check Point 1Check Point 1

Solve: .843 x

SOLUTIONSOLUTION

1) Isolate a radical on one side.

Simplify.

.843 x

2) Raise both sides to the nth power. Because n, the index, is 2, we square both sides.

22843 x

6443 x



This is the equation from step 2.

3) Solve the resulting equation.

6443 x

CONTINUECONTINUEDD

Subtract 5 from both sides.603 xDivide both sides by 2.20x

4) Check the proposed solution in the original equation.

Check 10:

843 x

8 4203

8460

864

88 true

The solution is 20.

The solution set is {20}.?

?

?



Number 4Number 4

Solve: .0523 x

SOLUTIONSOLUTION

1) Isolate a radical on one side. The radical can be isolated by adding 5 to both sides. We obtain

Simplify.

.523 x

2) Raise both sides to the nth power.

2523 x

2523 x




CONTINUECONTINUEDD

273 x9x


Check 9:

0523 x

0 5-293

0525

00 true

The solution is 9.

The solution set is {9}.?

?

2523 x



EXAMPLE (Also number 6)EXAMPLE (Also number 6)

Solve: .61152 x

SOLUTIONSOLUTION

1) Isolate a radical on one side. The radical, , can be isolated by subtracting 11 from both sides. We obtain

Simplify.

52 x

.552 x


22552 x

2552 x



This is the equation from step 2.


2552 x

CONTINUECONTINUEDD

Subtract 5 from both sides.202 xDivide both sides by 2.10x


Check 10:

61152 x

6115102

61125

6115

616 false

Therefore there is no solution to the equation.

?

?

?



EXAMPLEEXAMPLE

Solve: .5733 xx

SOLUTIONSOLUTION

1) Isolate a radical on one side. The radical, , can be isolated by subtracting 3x from both sides. We obtain

Simplify. Use the special formula

73 x

.3573 xx


29302573 xxx

223573 xx

.2 222 BABABA



Equation from step 2.

3) Solve the resulting equation. Because of the -term, the resulting equation is a quadratic equation. We need to write this quadratic equation in standard form. We can obtain zero on the left side by subtracting 3x and 7 from both sides.

2530973 2 xxx

CONTINUECONTINUEDD 2x

Subtract 3x and 7 from both sides.

182790 2 xx

Factor out the GCF, 9. 2390 2 xx

Divide both sides by 9.230 2 xx

Factor the right side. 120 xx



CONTINUECONTINUEDD

Set each factor equal to 0.02 x 01xSolve for x.2x 1x

4) Check the proposed solutions in the original equation.

Check -2: Check -1:5733 xx 5733 xx

572323 571313

516 543

516 523 57 55

The solution is -1. The solution set is {-1}.

truefalse

?

?

? ?

?

?



Number 10Number 10

Solve: .712 xx

SOLUTIONSOLUTION


Check 12:

712 xx

7121122

525

55 true

?

?

12x

Do on board


In Summary…

Solving Radical Equations Containing nth Roots

1.Isolate one radical on one side of the equation.2.Raise both sides to the nth power3.Solve the resulting equation4.Check proposed solutions in the original equation.

Sometimes proposed solutions will work in the final simplified form of the original equation, but will not work in the original equation itself. These imposter roots that sometimes slip in when we square both sides of an equation are called extraneous roots.

DONE



EXAMPLEEXAMPLE

Solve: .444 xx

SOLUTIONSOLUTION

1) Isolate a radical on one side. The radical, , can be isolated by subtracting from both sides. We obtain

Simplify. Use the special formula

4x


448164 xxx .2 222 BABABA

4x

.444 xx

22444 xx



Combine like terms.48204 xxxCONTINUECONTINUE

DD

1) Isolate a radical on one side. The radical, can be isolated by subtracting 20 + x from both sides and then dividing both sides by -8. We obtain

,4x

.43 x


22 43 x

49 x Square the 3 and the .4x



CONTINUECONTINUEDD


49 xThis is the equation from the last step.

x5 Subtract 4 from both sides.


444 xx

Check 5:

44545

491

431

44

The solution is 5. The solution set is {5}.

true

?

?

?



EXAMPLEEXAMPLE

Solve: .10732 4

1

x

SOLUTIONSOLUTION

Although we can rewrite the equation in radical form

This is the given equation.

it is not necessary to do so. Because, the equation involves a fourth root, we isolate the radical term – that is, the term with the rational exponent – and raise both sides to the 4th power.

,107324 x

10732 4

1

x

Subtract 7 from both sides. 332 4

1

x



Raise both sides to the 4th power. 44

4

1

332

x

CONTINUECONTINUEDD

Multiply exponents on the left sides and then simplify.

8132 x

Subtract 3 from both sides.782 x

Divide both sides by 2.39x

Upon checking the proposed solution, 39, in the original equation, we find that it checks out and is a solution. Therefore the solution set is {39}.



EXAMPLEEXAMPLE

For each planet in our solar system, its year is the time it takes the planet to revolve once around the sun. The function

models the number of Earth days in a planet’s year, f (x), where x is the average distance of the planet from the sun, in millions of kilometers. Use the function to solve the following problem.

There are approximately 88 Earth days in the year of the planet Mercury. What is the average distance of Mercury from the sun? Use a calculator and round to the nearest million kilometers.

2

3

2.0 xxf



SOLUTIONSOLUTION

CONTINUECONTINUEDD

To find the average distance of Mercury from the sun, replace f (x) in the function with 88.

2

3

2.0 xxf This is the given equation.

2

3

2.088 x Replace f (x) with 88.

2

3

440 x Divide both sides by 0.2.

22/32440 x Square both sides.

3600,193 x Simplify.



CONTINUECONTINUEDD 3 33 600,193 x Take the cube root of both sides.

x58 Simplify.

The model indicates that the average distance, to the nearest million kilometers, that Mercury is from the sun is 58 million kilometers.

DONE