EXTRANEOUS SOLUTIONS OF RADICAL EQUATIONS

Example 1 :

Solve √(x-2)  =  (x- 4)

Solution :

[√(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.

Example 2 :

Solve x  =  √(6 - x)

Solution :

x  =  √(6 - x)

Take squares on 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.

Example 3 :

Solve x  =  √(x + 20)

Solution :

x  =  √x + 20

Take squares on 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.

Example 4 :

If an object is dropped from an initial height h its velocity at impact with the ground is given by

v = √2gh

where g is the acceleration due to gravity and h is the initial height.

a) Find the initial height (in feet) of an object of its velocity at impact is 44 ft/sec (Assume the acceleration due to gravity is g = 32 ft/sec²)

b)  Find the initial height (in meters) of an object of its velocity at impact is 26 ft/sec (Assume the acceleration due to gravity is g = 9.8 ft/sec²)

Solution :

v = √2gh

a)  h = ?, v = 44 ft/sec and g = 32 ft/sec²

44 = √2(32) h

Taking square on both sides, 

44² = 2(32) h

h = 1936/64

h = 30.25 feet

b) h = ?, v = 26 ft/sec and g = 9.8 ft/sec²

26 = √2(9.8) h

Taking square on both sides, 

26² = 2(9.8) h

h = 676/19.6

h = 34.48 feet

Example 5 :

Solve for x

√(3 x - 5) - √3x = -1

Solution :

√(3 x - 5) - √3x = -1

√(3 x - 5) = -1 + √3x

√(3 x - 5) = √3x - 1

Squaring on both sides, we get

3x - 5 = (√3x - 1)²

3x - 5 = (√3x)² - 2√3x(1) + 1²

3x - 5 = 3x - 2√3x + 1

-5 = - 2√3x + 1

-5 - 1 = -2√3x

-6 / (-2) = √3x

3 = √3x

3² = 3x

9 = 3x

x = 9/3

x = 3

Example 6 :

Solve for x

6 = √(x² + 3) - x

Solution :

6 = √(x² + 3) - x

6 + x = √(x² + 3)

Squaring on both sides, we get

(6 + x)² = x² + 3

6² + 2(6)x + x² = x² + 3

36 + 12x + x² = x² + 3

36 + 12x = 3

12x = 3 - 36

12x = -33

x = -33/12

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