SOLVE THE EQUATION ELIMINATING ANY EXTRANEOUS SOLUTIONS

Example 1 :

Solve for x :

2x/(x - 1) + 1/(x - 3)  =   2/x² - 4x + 3

Solution :

2x/(x - 1) + 1/(x - 3)  =   2/x² - 4x + 3

Factoring the quadratic equation x² - 4x + 3, we get factors as (x - 1) (x - 3).

2x/(x - 1) + 1/(x - 3)  =   2/(x - 1) (x - 3)  --(1)

2x(x - 3) + 1(x - 1)  =  2

2x² - 6x + x - 1  =  2

2x² - 5x - 1 - 2  =  0

2x² - 5x - 3  =  0

(2x + 1) (x - 3)  =  0

To check whether above values are solution or not, we apply the values in (1).

If x  =  -1/2, we get the defined value.

If x  =  3, we get undefined value. So, only x  =  -1/2 is a solution.

Example 2 :

Solve for x :

(x - 3)/x  +  3/(x + 2)  +  6/(x² + 2x)  =  0

Solution :

(x - 3)/x  +  3/(x + 2)  +  6/(x² + 2x)  =  0

(x - 3)/x  +  3/(x + 2)  +  6/x(x + 2)  =  0

Least common multiple is x (x + 2).

(x - 3)(x + 2) + 3x  +  6  =  0

x² - x - 6 + 3x + 6  =  0

x² + 2x  =  0

x(x + 2)  =  0

x  =  0  and x + 2  =  0

x  =  0 and x  =  -2

Substituting x  =  0 or x  =  -2 into the original equation results in the division is zero. So both of these numbers are extraneous solution. So the original equation has no solution.

Example 3 :

Solve algebraically for x :

√(x + 5) + 1 = x

Solution :

√(x + 5) + 1 = x

√(x + 5) = x - 1

Squaring on both sides

√(x + 5)² = (x - 1)²

x + 5 = x² - 2x(1) + 1²

x + 5 = x² - 2x + 1

x² - 2x - x - 5 + 1 = 0

x² - 3x - 4 = 0

x² - 4x + 1x - 4 = 0

x(x - 4) + 1(x - 4) = 0

(x - 4)(x + 1) = 0

x = 4 and x = 1

Example 4 :

To solve the equation

7/(x + 7) + 4x/(x - 7) = (3x + 7)/(x - 7)

Joan's first step to multiply both sides by the least common denominator. Which statement is true?

a)  -14 is an extraneous solution 

b)  7 is an extraneous solution 

c) 7 and -7 are extraneous solutions.

d) There are no extraneous solutions.

Solution :

7/(x + 7) + 4x/(x - 7) = (3x + 7)/(x - 7)

Least common multiple of (x + 7) and (x - 7) is (x + 7)(x - 7).

[7(x - 7) + 4x(x + 7)]/(x + 7) (x - 7) = (3x + 7)/(x - 7)

(7x - 49 + 4x² + 28x) /(x + 7) (x - 7) = (3x + 7)/(x - 7)

(4x² + 28x + 7x - 49) / (x + 7) = 3x + 7

(4x² + 35x - 49) = (3x + 7)(x + 7)

4x² + 35x - 49 = 3x² + 21x + 7x + 49

4x² + 35x - 49 = 3x² + 28x + 49

4x² - 3x² + 35x - 28x - 49 - 49 = 0

x² + 7x - 98 = 0

x² + 14x - 7x - 98 = 0

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

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

x = 7 and x = -14

So, 7 is an extraneous solution.

Example 5 :

To solve

2x/(x - 2) - 11/x = 8/(x² - 2x)

Ren multiplied both side by the least common denominator. Which statement is true ?

1) 2 is an extraneous solution.

2) 0 and 2 are extraneous solutions.

3) 7/2 is an extraneous so1ution.

4) This equation does not contain any extraneous solutions

Solution :

2x/(x - 2) - 11/x = 8/(x² - 2x)

2x/(x - 2) - 11/x = 8/x(x - 2)

[2x(x) - 11(x - 2)] / x(x - 2) = 8/x(x - 2)

2x² - 11(x - 2) = 8

2x² - 11x + 22 = 8

2x² - 11x + 22 - 8 = 0

2x² - 11x + 14 = 0

2x² - 4x - 7x + 14 = 0

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

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

x = 7/2 and x = 2

2 is an extraneous solution. So, option a is correct.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.