RADICAL EQUATIONS WITH EXTRANEOUS SOLUTIONS WORKSHEET

Problem 1 :

Solve √(x - 2) = (x - 4)

Solution :

√(x-2)  =  (x- 4)

Squaring both sides,

[√(x-2)]²  =  (x- 4)²

(x - 2)  =  (x² - 2 x (4) + 4²)

x - 2  =  x² - 8 x + 16

x² - 8 x + 16 - x + 2  =  x - 2 - x + 2

x² - 9 x + 18  =  0

(x - 6) (x - 3)  =  0

let us apply the above values one by one to check which is an extraneous solution.

Since 3 does not satisfy the original equation, 6 is the only solution.

Hence 3 is the extraneous solution and 6 is the solution.

Problem 2 :

Solve for x

x  =  √(6 - x)

Solution :

x  =  √(6 - x)

Squaring both sides,

x²  =  [√(6 - x)]²

x²  =  6 - x 

Add x and subtract 6 on both sides, we get

x² - 6 + x  =  6 - x + x - 6

x² + x - 6  =  0

By factoring the above quadratic equation, we get

(x - 2)(x + 3)  =  0

Now let us apply each values one by one in the original equation,

Since -3 does not satisfy the original equation, 2 is the only solution.

Hence 2 is the extraneous solution and 6 is the solution.

Problem 3 :

Solve for x

x  =  √x + 20

Solution :

x  =  √x + 20

Squaring both sides,

x²  =  [√(x + 20)]²

x²  =  x + 20 

x²  - x - 20  =  x + 20 - x - 20

x²  - x - 20  =  0

By factoring the above quadratic equation, we get

(x - 5)(x + 4)  =  0

Now let us apply each values one by one in the original equation,

Since -4 does not satisfy the original equation, 5 is the only solution.

Hence -4 is the extraneous solution and 5 is the solution.

Problem 4 :

The number of dogs, D, housed at a county animal shelter is modeled by the function

D = 4√(2M) + 50

where M is the number of months the shelter has been open. How many months will it take for 74 dogs to be housed at the shelter?

Solution :

D = 4√(2M) + 50

Given that, number of dogs = 74

74 = 4√(2M) + 50

74 - 50 = 4√(2M)

24 = 4√(2M)

24/4 = √(2M)

6 = √(2M)

Squaring on both sides

62 = 2M

36 = 2M

M = 36/2

M = 18

Problem 5 :

What is the solution set of the equation 

y = 2 + √(y² - 12)

Solution :

y = 2 + √(y² - 12)

y - 2 = √(y² - 12)

(y - 2)² = y² - 12

y² - 2(y)(2) + 2² = y² - 12

y² - 4y + 4 = y² - 12

-4y + 4 = -12

-4y = -12 - 4

-4y = -16

y = 16/4

y = 4

So, the value of y is 4.

Problem 6 :

Solve algebraically for x 

√(x² + x - 1) + 11x = 7x + 3

Solution :

√(x² + x - 1) + 11x = 7x + 3

√(x² + x - 1) = 7x - 11x + 3

√(x² + x - 1) =  -4x + 3

(x² + x - 1) =  (-4x + 3)²

(x² + x - 1) = (3 - 4x)²

(x² + x - 1) = 3² - 2(3)(4x) + (4x)²

(x² + x - 1) = 9 - 24x + 16x²

16x² - x² - 24x - x + 9 + 1 = 0

15x² - 25x + 10 = 0

3x² - 5x + 2 = 0

3x² - 2x - 3x + 2 = 0

x(3x - 2) - 1(3x - 2) = 0

(x - 1)(3x - 2) = 0

x = 1 and x = 2/3

So, the values of x are 1 and 2/3.

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