# SOLVING RADICAL EQUATIONS EXAMPLES

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.

## Solving radical equations examples

Example 1 :

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.

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  =  42 - 2 x (4) + x2

6 - x  =  16 - 8 x + x2

Subtract 6 and add x on both sides, we get

6 - x + x - 6   =  16 - 8 x + x2 + x - 6

0  =  10 - 7 x + x2

x2 - 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)  =  4x  =  55 +  √(6 - 5)  =  45 +  1  =  46 ≠  4 x + √(6 - x)  =  4x  =  22 +  √(6 - 2)  =  42 +  2  =  44 =  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  =  32

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  =  x2 - 2x(4) + 42

x + 2  =  x2 - 8x + 16

Subtract x and 2 on both sides,

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

0  =  x2 - 9x + 14

x2 - 9x + 14  =  0

Factoring the above quadratic equation,

(x - 2) (x - 7)  =  0

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

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

 √(x + 2)  =  x - 4√(2 + 2)  =  2 - 4√4  =  - 22  ≠  -2 √(x + 2)  =  x - 4√(7 + 2)  =  7 - 4√9  =  33  =  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  =  x2 - 2x(5) + 52

3x - 5  =  x2 - 10x + 25

Subtract 3x and add 5 on both sides,

3x - 5 - 3x + 5  =  x2 - 10x + 25 - 3x + 5

0  =  x2 - 13x + 30

x2 - 13x + 30  =  0

Factoring the above quadratic equation,

(x - 10) (x - 3)  =  0

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

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

 √(3x - 5)  =  x - 5x = 10 √30 - 5  =  10 - 5√25  =  55  =  5 √(3x - 5)  =  x - 5x = 3 √9 - 5  =  3 - 5√4  =  -22  ≠  -2

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

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