how do we solve radical equations? do now: simplify the given expression. do now: simplify the given...

How Do We Solve Radical How Do We Solve Radical Equations?Equations?

• Do Now: Do Now: Simplify the given Simplify the given expression.expression.

1. 2. 1. 2. 

An equation in which a variable occurs in the radicandis called a radical equation. It should be noted, that when solving a radical equation algebraically, extraneous roots may be introduced when both sides ofan equation are squared. Therefore, you must check your solutions for a radical equation.

Solve: √ x - 3 - 3 = 0

√ x - 3 = 3

(√ x - 3 )2 = (3)2

x - 3 = 9 x = 12

Check:

√ x - 3 - 3

√ 12 - 3 - 33 - 3 0

0

Therefore, the solution is x = 12.

x ≥ 3

Radical Equations

L.S. R.S.

If

,5212 x

then x is equal to (1) 1 (3) 5 (2) 2 (4) 4

5

102

912

312

5212

x

x

x

x

x

What is the solution of the equation

?6332 x

(1) 42 (3) 3 (2) 39 (4) 6

42

842

8132

932

6332

x

x

x

x

x

4 + √ 4 + x2 = x

√ 4 + x2 = x - 4

4 + x2 = x2 - 8x + 16 8x = 12

x

3

2

3

2

13

2,Since the solution of

x = 3

2 is extraneous. Therefore,

there are no real roots.

Check: 4 4 x2

4 4 3

2

2

4 4 9

4

4 25

4

4 5

213

23

2≠

(√ 4 + x2)2 = (x - 4)2

x 3

2

Solving Radical Equations

The solution set of the equation

x x 6

is

(1) {–2,3} (3) {3} (2) {–2} (4) { }

x x

x x

x x

x x

x x

6

6

6 0

3 2 0

3 2

2

2

( )( )

What is the solution set of the equation

9 10x x ? (1) {-1} (3) {10} (2) {9} (4) {10, -1}

9 10

9 10

9 10 0

10 1 0

10 1

2

2

x x

x x

x x

x x

x x

( )( )

x = -1 is an extraneous solution.

2x 4 x 7 0.

Set up the equation so thatthere will be one radical oneach side of the equal sign.

2x 4 x 7

Square both sides.2x 4 2 x 7 2

Simplify. 2x + 4 = x + 7 x = 3

Verify your solution.

2x 4 x 7 0Therefore, thesolution isx = 3.

x ≥ -2Solve

Solving Radical Equations

2(3) 4 3 7

10 10

0

L.S. R.S.

Squaring a Binomial

(a + 2)2 = a2 + 4a + 4Note that the middle term is twice the product of the two terms of the binomial.

(a√x + b)2

( 5 + √x - 2 )2

The middle term will be twice the product of the two terms.

5 x 2 2

10 x 2

5 x 2 2 A final concept that you should know:

25 10 x 2 (x 2)

x 23 10 x 2 = a2x + ab= a2(x + b)

5x 1 3x 5 2.Set up the equationso that there will beonly one radical oneach side of the equal sign.

Square both sidesof the equation.

Simplify.

Simplify by dividingby a common factor of 2.

Square both sides of the equation.

5x 1 2 2 3x 5 2

5x 1

5x 1 3x 1 4 3x 5

2x 2 4 3x 5

x 1 2 3x 5

x 1 2 2 3x 5 2x2 2x 1 4(3x 5) Use Foil.

5x 1 2 3x 5

Use Foil.4 4 3x 5 (3x 5)

Solve

Solving Radical Equations

x2 2x 1 4(3x 5) Distribute the 4.

x2 2x 1 12x 20 Simplify.

x2 10x 21 0 Factor the quadratic.

(x 3)(x 7) 0Solve for x.

x - 3 = 0 or x - 7 = 0 x = 3 or x = 7 Verify both solutions.

5x 1 3x 5 2

5(3) 1 3(3) 5

4 2

2

5x 1 3x 5 2

5(7) 1 3(7) 5

6 4

2

Solving Radical Equations

L.S. R.S. L.S. R.S.

One more to see another extraneous solution:

313 xx The radical is already isolated

2 2You must square the whole side

NOT each term.

9613 2 xxx

Square both sides

Since you have a quadratic equation (has an x2 term) get everything on one side = 0 and see if you can factor this

1,8 xx

You MUST check these answers

55

38183

313 xxThis must be FOILed

0892 xx

018 xx 22

31113

Doesn't work!Extraneous

It checks!

a solution that you find algebraically but DOES NOT make a true statement when you substitute it back into the equation.

Let's try another one:

0112 3

1

x First isolate the radical

- 1 - 1

112 3

1

x3 3

112 x

Now since it is a 1/3 power this means the same as a cube root so cube both sides

Now solve for x

- 1 - 1

22 x1x

Let's check this answer

011123 00 It checks!

Remember that the 1/3 power means the same thing as a cube root.

y x 2.Graph

The domain is x > -2.The range is y > 0.

Graphing a Radical Function

Solve x 3 3 0. The solution will be theintersection of the graph

y x 3 3

and the graph ofy = 0.

The solutionis x = 12.

Check:

x 3 3 0

12 3 39 33 3

Solving a Radical Equation Graphically

L.S. R.S.

5x 1 3x 5 2.

The solution isx = 3 or x = 7.

Solve

Solving a Radical Equation Graphically

7x 3 2 3.Find the values for which the graph of

y 7x 3 2

is above the graph of y = 3.

The graphs intersect at x = 4.

Note the radical7x - 3 is defined only

when .

Therefore, the solutionis x > 4.

x > 4

Solving Radical Inequalities

Solve

x 3

7

x 1 3.

The solution is -1 < x and x < 8.

The graphs intersectat the point where x = 8.

x ≥ -1 and x < 8

x > -1

Solving Radical Inequalities

Solve

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