SOLVING SQUARE ROOT EQUATIONS

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)]² = (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)]² = 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)]² = [√(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)]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]² = [3√y]²

x = 3²(√y)²

x = 9y

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