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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

5. Radical Equations

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Exercise 37 Page 262

What are some of the steps you take when rewriting a radical equation as an equation without radicals?

See solution.

Practice makes perfect

Each time we solve a radical equation, we follow a few steps.

Isolate the radical term on one side of the equation by adding or subtracting terms from both sides of the equation.
If the radical is multiplied by any factor other than 1, divide the equation by this factor.
Eliminate the radical by raising both sides of the equation to the power of the radical's index.
Since the equation no longer contains a radical, solve it using appropriate methods.
Check if any of the answers is an extraneous solution.

Notice, that sometimes one or more steps can be skipped. For example when the radical term is already isolated — we start with the second step. We will demonstrate all these steps in an example equation.

Solving an Equation

Consider the following radical equation. - 2x +11 + 2sqrt(x+2)= 3 We will solve this equation by going through each of the steps that we previously listed. As such, we will start by isolating the radical term using the Properties of Equality. Then we will raise the equation to the second power in order to eliminate the square root.

- 2x +11 + 2sqrt(x+2)= 3

▼

Solve for sqrt(x+2)

LHS-11=RHS-11

- 2x + 2sqrt(x+2)= - 8

LHS+2x=RHS+2x

2sqrt(x+2)= - 8 + 2x

.LHS /2.=.RHS /2.

sqrt(x+2)= - 4 + x

CommutativePropAdd

Commutative Property of Addition

sqrt(x+2)= x-4

LHS^2=RHS^2

( sqrt(x+2) )^2 = (x-4)^2

( sqrt(a) )^2 = a

x+2 = (x-4)^2

Now, we will solve the equation by factoring and applying the Zero Product Property. Let's start by writing all the terms on one side of the equals sign.

x+2 = (x-4)^2

LHS-2=RHS-2

x = (x-4)^2 -2

LHS-x=RHS-x

0 = (x-4)^2 -2 -x

▼

Simplify

ExpandNegPerfectSquare

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

0 = x^2 -8x +16 -2 -x

Subtract terms

0=x^2-9x+14

Rearrange equation

x^2-9x+14=0

Write as a difference

x^2-7x -2x +14=0

▼

Factor

Factor out x

x(x-7)-2x +14=0

Factor out - 2

x(x-7)-2(x -7)=0

Factor out (x-7)

(x-7)(x-2)=0

Use the Zero Product Property

lcx-7=0 & (I) x-2=0 & (II)

▼

(I), (II): Solve for x

(I): LHS+7=RHS+7

lx=7 x-2=0

(II): LHS+2=RHS+2

lx_1=7 x_2=2

Finally, we need to check the answers to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=7 and x_2=2 one at a time. Let's start by substituting x= 7 into the original equation.

- 2x +11 + 2sqrt(x+2)= 3

x= 7

- 2( 7) +11 + 2sqrt(7+2)? =3

▼

Simplify

Add terms

- 2(7) +11 + 2sqrt(9)? =3

Calculate root

- 2(7) +11 + 2(3)? =3

Multiply

- 14 +11 + 6? =3

Add terms

3=3 ✓

We got a true statement, so x=7 is a valid solution. Now, we can substitute x= 2.

- 2x +11 + 2sqrt(x+2)= 3

x= 2

- 2( 2) +11 + 2sqrt(2+2)? =3

▼

Simplify

Add terms

- 2(2) +11 + 2sqrt(4)? =3

Calculate root

- 2(2) +11 + 2(2)? =3

Multiply

- 4 +11 + 4? =3

Add terms

11 ≠ 3 *

In this case, we got a false statement. Therefore, x=7 is the only solution to the original equation.

Subchapter links

Exercises p.261-263