Extraneous solutions of radical equations :

Squaring each side of an equation sometimes produces extraneous solutions. An extraneous solution is a solution derived from an equation that is not a solution of the original equation.

Therefore, you must check all solutions in the original equation when you solve radical equations.

Example 1 :

Solve √(x-2)  =  (x- 4)

Solution :

[√(x-2)]2  =  (x- 4)2

(x - 2)  =  (x- 2 x (4) + 42)

x - 2  =  x- 8 x + 16

x- 8 x + 16 - x + 2  =  x - 2 - x + 2

x- 9 x + 18  =  0

(x - 6) (x - 3)  =  0

 x - 6  =  0Add 6 on both sidesx  =  6 x - 3  =  0Add 3 on both sidesx  =  3

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

 x  =  6√(6-2)  =  (6-4)√4  =  22  =  2 x  =  3√(3-2)  =  (3 - 4)√1  = -11  ≠  -1

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

 x - 2  =  0Add 2 on both sides, we getx - 2 + 2  =  0 + 2x  =  2 x + 3  =  0Subtract 3 on both sides, we getx + 3 - 3  =  0 - 3x  =  -3

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

 x  =  √(6 - x)x  =  22  =  √(6 - 2)2  =  √42  =  2 x  =  √(6 - x)x  =  -3-3  =  √(-3 - 2)-3 ≠  √-5

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

 x - 5  =  0Add 5 on both sidesx - 5 + 5  =  0 + 5x  =  5 x + 4  =  0Subtract 4 on both sidesx + 4 - 4  =  0 - 4x  =  -4

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

 x  =  √(x + 20)x  =  55  =  √(5 + 20)5  =  √255  =  5 x  =  √(x + 20)x  =  -4-4  =  √(-4 + 20)-4  =  √16-4 ≠ 4

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

Hence -4 is the extraneous solution and 5 is the solution. After having gone through the stuff given above, we hope that the students would have understood, "Extraneous solutions of radical equations".

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