**Solving radical equations examples :**

To solve these equations, first isolate the radical on one side of the equation. Then square each side of the equation to eliminate the radical.

**Example 1 :**

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.

Let us look into the next example on "Solving radical equations examples".

**Example 2 :**

Solve x + √(6 - x) = 4

**Solution :**

x + √(6 - x) = 4

Subtract x on both sides,

x + √(6 - x) - x = 4 - x

√(6 - x) = 4 - x

Take squares on both sides

[√(6 - x)]^{2} = (4 - x)^{2}

6 - x = 4^{2} - 2 x (4) + x^{2}

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

Subtract 6 and add x on both sides, we get

6 - x + x - 6 = 16 - 8 x + x^{2} + x - 6

0 = 10 - 7 x + x^{2}

x^{2} - 7 x + 10 = 0

(x - 5) (x - 2) = 0

x = 5 and x = 2

By applying the values one by one in the equation, we get

x + √(6 - x) = 4 x = 5 5 + √(6 - 5) = 4 5 + 1 = 4 6 ≠ 4 |
x + √(6 - x) = 4 x = 2 2 + √(6 - 2) = 4 2 + 2 = 4 4 = 4 |

Since 5 does not satisfy the original equation, 2 is the only solution.

Let us look into the next example on "Solving radical equations examples".

**Example 3 :**

Solve √(x + 1) + 7 = 10

**Solution :**

√(x + 1) + 7 = 10

Subtract 7 on both sides,

√(x + 1) + 7 - 7 = 10 - 7

√(x + 1) = 3

Take squares on both sides,

[√(x + 1)]^{2} = 3^{2}

x + 1 = 9

Now subtract 1 on both sides, we get

x + 1 - 1 = 9 - 1

x = 8

Hence the value of x is 8.

Let us look into the next example on "Solving radical equations examples".

**Example 4 :**

Solve √(x + 2) = x - 4

**Solution :**

√(x + 2) = x - 4

Take squares on both sides,

[√(x + 2)]^{2} = (x - 4)^{2}

We can expand the expression (x - 4)^{2} using the algebraic identities (a - b)^{2}

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

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

Subtract x and 2 on both sides,

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

0 = x^{2} - 9x + 14

x^{2} - 9x + 14 = 0

Factoring the above quadratic equation,

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

(x - 2) = 0 Add 2 on both sides, x - 2 + 2 = 0 + 2 x = 2 |
(x - 7) = 0 Add 7 on both sides, x - 7 + 7 = 0 + 7 x = 7 |

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

√(x + 2) = x - 4 √(2 + 2) = 2 - 4 √4 = - 2 2 ≠ -2 |
√(x + 2) = x - 4 √(7 + 2) = 7 - 4 √9 = 3 3 = 3 |

Since 2 does not satisfy the original equation, 7 is the only solution.

**Example 5 :**

Solve √(3x - 5) = x - 5

**Solution :**

√(3x - 5) = x - 5

Take squares on both sides,

[√(3x - 5)]^{2} = (x - 5)^{2}

We can expand the expression (x - 5)^{2} using the algebraic identities (a - b)^{2}

3x - 5 = x^{2} - 2x(5) + 5^{2}

3x - 5 = x^{2} - 10x + 25

Subtract 3x and add 5 on both sides,

3x - 5 - 3x + 5 = x^{2} - 10x + 25 - 3x + 5

0 = x^{2} - 13x + 30

x^{2} - 13x + 30 = 0

Factoring the above quadratic equation,

(x - 10) (x - 3) = 0

(x - 10) = 0 Add 10 on both sides, x - 10 + 10 = 0 + 10 x = 10 |
(x - 3) = 0 Add 3 on both sides, x - 3 + 3 = 0 + 3 x = 3 |

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

√(3x - 5) = x - 5 x = 10 √30 - 5 = 10 - 5 √25 = 5 5 = 5 |
√(3x - 5) = x - 5 x = 3 √9 - 5 = 3 - 5 √4 = -2 2 ≠ -2 |

Since 3 does not satisfy the original equation, 10 is the only solution.

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