SOLVING RADICAL EQUATIONS WITH SQUARE ROOTS

Solve for x :

Example 1 :

5√2 = √x

Solution :

To remove the square root, we can take squares on both sides.

(5√2)² = (√x)²

Distributing the power for all the terms that we have in the power.

5²√2² = (√x)²

25 (2) = x

x = 50

Example 2 :

3√x = √45

Solution :

To remove the square root, we can take squares on both sides.

(3√x)² = (√45)²

Distributing the power for all the terms that we have in the power.

3²√x² = (√45)²

9x = 45

Divide by 9 on both sides.

x = 45/9

x = 5

Example 3 :

2√2 = √4x

Solution :

To remove the square root, we can take squares on both sides.

(2√2)² = (√4x)²

Distributing the power for all the terms that we have in the power.

2²√2² = (√4x)²

4(2) = 4x

Divide by 4 on both sides.

x = 8/4

x = 2

Example 4 :

√(x+1) = 6

what value of x satisfies the equation above ?

(a) 5   (b) 6   (c)  35   (d)  36

Solution :

√(x+1) = 6

Taking squares on both sides.

x+1 = 6²

x+1 = 36

Subtracting 1 on both sides.

x = 36-1

x = 35

Example 5 :

(x-12) = √(x+44)

What are all possible solution to the given equation ?

(a)  5   (b)  20   (c)  -5 and 20   (d)  5 and 20

Solution :

(x-12) = √(x+44)

Take squares on both sides.

(x-12)² = (x+44)

Using the algebraic identity,

(a-b)² = a²-2ab+b²

x²-2x(12)+12² = x+44

x²-24x+144 = x+44

Subtract x and 44 on both sides.

x²-25x+100 = 0

(x-20)(x-5) = 0

Equating each factor to zero, we get

x = 20 and x = 5

Example 6 :

x + 3√x = 28

Solution :

x + 3√x = 28

Subtract x on both sides.

3√x = 28 - x

Take squares on both sides.

(3√x)² = (28 - x)²

9x = 784 - 56x + x²

By subtracting 9x on both sides.

x² - 65x+784 = 0

(x-49)(x-16) = 0

x = 49 and x = 16