# SOLVING RADICAL EQUATIONS WORKSHEET

Solving radical equations worksheet :

Here we are going to see some practice questions on solving radical equations worksheet.

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 worksheet-Practice questions

(1)  Solve x  =  √(6 - x)

(2)  Solve x + √(6 - x)  =  4

(3)  Solve √(x + 1) + 7  =  10

(4)  Solve √(x + 2)  =  x - 4

(5)  Solve √(3x - 5)  =  x - 5

(6)  Solve √(7x+18)  =  9

(7)  Solve √(x2 + 9x + 14) =  x + 4

## Solving radical equations worksheet-Solution

Question 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 solution of next problem on "Solving radical equations worksheet".

Question 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 solution of next problem on "Solving radical equations worksheet".

Question 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.

Question 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.

Question 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.

Question 6 :

Solve √(7x+18)  =  9

Solution :

[√(7x + 18)]2  =  (9)2

7x + 18  =  81

Subtract 18 on both sides

7x + 18 - 18  =  81 - 18

7x  =  63

Divide by 7 on both sides

7x/7  =  63/7

x  =  9

Question 7 :

Solve √(x2 + 9x + 14) =  x + 4

Solution :

[√(x2 + 9x + 14)]2  =  (x + 4)2

x2 + 9x + 14  =  x+ 2x (4) + 42

x2 + 9x + 14  =  x+ 8x + 16

x2 + 9x + 14 - x- 8x - 16  =  0

x - 2  =  0

Add 2 on both sides

x - 2 + 2   =  0 + 2

x  =  2

After having gone through the stuff given above, we hope that the students would have understood "Solving radical equations worksheet".

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