**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 }- 2 x (4) + 4^{2})

x - 2 = x^{2 }- 8 x + 16

x^{2 }- 8 x + 16 - x + 2 = x - 2 - x + 2

x^{2 }- 9 x + 18 = 0

(x - 6) (x - 3) = 0

x - 6 = 0 Add 6 on both sides x = 6 |
x - 3 = 0 Add 3 on both sides x = 3 |

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

x = 6 √(6-2) = (6-4) √4 = 2 2 = 2 |
x = 3 √(3-2) = (3 - 4) √1 = -1 1 ≠ -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,

x^{2} = [√(6 - x)]^{2}

x^{2} = 6 - x

Add x and subtract 6 on both sides, we get

x^{2} - 6 + x = 6 - x + x - 6

x^{2} + x - 6 = 0

By factoring the above quadratic equation, we get

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

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

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

x = √(6 - x) x = 2 2 = √(6 - 2) 2 = √4 2 = 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,

x^{2} = [√(x + 20)]^{2}

x^{2} = x + 20

x^{2} - x - 20 = x + 20 - x - 20

x^{2} - x - 20 = 0

By factoring the above quadratic equation, we get

(x - 5)(x + 4) = 0

x - 5 = 0 Add 5 on both sides x - 5 + 5 = 0 + 5 x = 5 |
x + 4 = 0 Subtract 4 on both sides x + 4 - 4 = 0 - 4 x = -4 |

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

x = √(x + 20) x = 5 5 = √(5 + 20) 5 = √25 5 = 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.

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