SOLVING SQUARE ROOT EQUATIONS

Key Concept :

When you solve square root equation, your first aim is to get rid of the square root. To get rid of the square root, you have to square both sides of the equation. Once you get rid of the square root, the equation can be solved easily.

In each case, solve for x.

Example 1 :

√x = 2

Solution :

√x = 2

Square both sides.

(√x)2 = 22

x = 4

Example 2 :

√x = -5

Solution :

√x = -5

Square both sides.

(√x)2 = (-5)2

x = 25

Example 3 :

9√x = 27

Solution :

9√x = 27

Divide both sides by 9.

√x = 3 

Square both sides.

(√x)2 = 32

x = 9

Example 4 :

√(3x + 1) = 4

Solution :

√(3x + 1) = 4

Square both sides.

[√(3x + 1)]2 = 42

3x + 1 = 16

Subtract 1 from both sides.

3x = 15

Divide both sides by 3.

x = 5

Example 5 :

5 + √(5x - 19) = 6

Solution :

5 + √(5x - 19) = 6

Subtract 5 from both sides.

√(5x - 19) = 1

Square both sides.

[√(5x - 19)]2 = 12

5x - 19 = 1

Add 19 to both sides.

5x = 20

Divide both sides by 5.

x = 4

Example 6 :

8 - √(x + 1) = 7

Solution :

8 - √(x + 1) = 7

Subtract 8 from both sides.

-√(x + 1) = -1

Square both sides.

[-√(x + 1)]2 = (-1)2

x + 1 = 1

Subtract 1 from both sides.

x = 0

Example 7 :

x = √(5x - 6)

Solution :

x = √(5x - 6)

Square both sides.

x2 = [√(5x - 6)]2

x2 = 5x - 6

Subtract 5x from both sides.

x2 - 5x = -6

Add 6 to both sides.

x2 - 5x + 6 = 0

Factor and solve it.

x2 - 3x - 2x + 6 = 0

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

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

x - 3 = 0  or  x - 2 = 0

x = 3  or  x = 2

Example 8 :

2 - x + √(x + 4) = 0

Solution :

2 - x + √(x + 4) = 0

Subtract 2 from both sides.

-x + √(x + 4) = -2

Add x to both sides.

√(x + 4) = x - 2

Square both sides.

[√(x + 4)]= (x - 2)2

x + 4 = (x - 2)(x - 2)

x + 4 = x2 - 2x - 2x + 4

x + 4 = x2 - 4x + 4

Subtract x and 4 from both ides.

0 = x2 - 5x

or

x2 - 5x = 0

x(x - 5) = 0

x = 0  or  x - 5 = 0

x = 0  or  x = 5

Example 9 :

8 +3√(3x - 5) = 20

Solution :

8 +3√(3x - 5) = 20

Subtract 8 from both sides.

3√(3x - 5) = 12

Divide both sides by 3.

√(3x - 5) = 4

Square both sides.

[√(3x - 5)]= 42

3x - 5 = 16

Add 5 to both sides.

3x = 21

Divide both sides by 3.

x = 7

Example 10 :

√(4x - 3) = √(2x + 1)

Solution :

√(4x - 3) = √(2x + 1)

Square both sides.

[√(4x - 3)]= [√(2x + 1)]2

4x - 3 = 2x + 1

Subtract 2x from both sides.

2x - 3 = 1

Add 3 to both sides.

2x = 4

Divide both sides by 2.

x = 2

Example 11 :

3 + √(x - 2) = √(4x + 1)

Solution :

3 + √(x - 2) = √(4x + 1)

Square both sides.

[3 + √(x - 2)]= [√(4x + 1)]2

[3 + √(x - 2)][3 + √(x - 2)] = 4x + 1

9 + 3√(x - 2) + 3√(x - 2) + (x - 2) = 4x + 1

9 + 6√(x - 2) + x - 2 = 4x + 1

6√(x - 2) + x + 7 = 4x + 1

Subtract x from both sides.

6√(x - 2) + 7 = 3x + 1

Subtract 7 to from both sides.

6√(x - 2) = 3x - 6

6√(x - 2) = 3(x - 2)

Divide both sides by 3.

2√(x - 2) = x - 2

Square both sides.

[2√(x - 2)]2 = (x - 2)2

22[√(x - 2)]2 = (x - 2)(x - 2)

4(x - 2) = x2 - 2x - 2x + 4

4x - 8 = x2 - 4x + 4

Subtract 4x from both sides.

-8 = x2 - 8x + 4

Add 8 to both sides.

0 = x2 - 8x + 12

or

x2 - 8x + 12 = 0

Factor and solve.

(x - 2)(x - 6) = 0

x - 2 = 0  or  x - 6 = 0

x = 2  or  x = 6

Example 12 :

If √x + √y = 4√y, where x > 0 and y > 0, what is x in terms of y?

Solution :

√x + √y = 4√y

Subtract √y from both sides.

√x = 3√y

Square both sides.

[√x]= [3√y]2

x = 32(√y)2

x = 9y

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