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§ 7.6. Radical Equations. 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. - PowerPoint PPT Presentation
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§ 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.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.6
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.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 7.6
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.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 7.6
Solving Radical Equations
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
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 7.6
Solving Radical Equations
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}.?
?
?
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 7.6
Solving Radical Equations
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
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 7.6
Solving Radical Equations
3) Solve the resulting equation.
CONTINUECONTINUEDD
273 x9x
4) Check the proposed solution in the original equation.
Check 9:
0523 x
0 5-293
0525
00 true
The solution is 9.
The solution set is {9}.?
?
2523 x
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 7.6
Solving Radical Equations
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
2) Raise both sides to the nth power. Because n, the index, is 2, we square both sides.
22552 x
2552 x
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 7.6
Solving Radical Equations
This is the equation from step 2.
3) Solve the resulting equation.
2552 x
CONTINUECONTINUEDD
Subtract 5 from both sides.202 xDivide both sides by 2.10x
4) Check the proposed solution in the original equation.
Check 10:
61152 x
6115102
61125
6115
616 false
Therefore there is no solution to the equation.
?
?
?
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 7.6
Solving Radical Equations
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
2) Raise both sides to the nth power. Because n, the index, is 2, we square both sides.
29302573 xxx
223573 xx
.2 222 BABABA
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 7.6
Solving Radical Equations
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
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 7.6
Solving Radical Equations
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
?
?
? ?
?
?
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 7.6
Solving Radical Equations
Number 10Number 10
Solve: .712 xx
SOLUTIONSOLUTION
4) Check the proposed solution in the original equation.
Check 12:
712 xx
7121122
525
55 true
?
?
12x
Do on board
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 7.6
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
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 7.6
Solving Radical Equations
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
2) Raise both sides to the nth power. Because n, the index, is 2, we square both sides.
448164 xxx .2 222 BABABA
4x
.444 xx
22444 xx
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 7.6
Solving Radical Equations
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
2) Raise both sides to the nth power. Because n, the index, is 2, we square both sides.
22 43 x
49 x Square the 3 and the .4x
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 7.6
Solving Radical Equations
CONTINUECONTINUEDD
3) Solve the resulting equation.
49 xThis is the equation from the last step.
x5 Subtract 4 from both sides.
3) Check the proposed solution in the original equation.
444 xx
Check 5:
44545
491
431
44
The solution is 5. The solution set is {5}.
true
?
?
?
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 7.6
Solving Radical Equations
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
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 7.6
Solving Radical Equations
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}.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 7.6
Solving Radical Equations
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
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 7.6
Solving Radical Equations
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.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 7.6
Solving Radical Equations
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