SOLVING EQUATIONS USING SQUARE ROOT PROPERTY

The inverse operation of square is square root.

The square root property says that if x2 = k

then, x = √k

  • If k is positive, then x has two real answers.
  • If k is negative, then x has two imaginary

Solve the following by square rooting both sides.

Problem 1 :

x2 = 81

Solution :

x2 = 81

Using square root property.

x = √81

x = ±9

So, the solution is {-9, 9}.

Problem 2 :

x2 = 24

Solution :

x2 = 24

Using square root property.

x = √24

x = √(2 2 2 3)

x = ±2√6

So, the solution is {-2√62√6}.

Problem 3 :

x2 = -16

Solution :

x2 = -16

Using square root property.

x =√-16

So, there is no real root.

Problem 4 :

(x – 5)2 = 36

Solution :

(x – 5)2 = 36

Using square root property.

x - 5 = √36

x – 5 = ±6

x - 5 = 6

x = 6 + 5

x = 11

x - 5 = -6

x = -6 + 5

x = -1

So, the solution is {-1, 11}.

Problem 5 :

(x + 2)2 = 27

Solution :

(x + 2)2 = 27

Using square root property.

x + 2 = √27

x + 2 = √(9  3)

x + 2 = √(3  3  3)

x + 2 = ±3√3

x + 2 = 3√3

x = 3√3 - 2

x + 2 = -3√3

x = -3√3 - 2

So, the solution is {3√3 - 2,  -3√3 - 2}

Problem 6 :

(x – 4)2 = -25

Solution :

(x – 4)2 = -25

Using square root property.

x - 4 = √-25

So, there is no real root.

Problem 7 :

2(x + 3)2 – 18 = 0

Solution :

2(x + 3)2 – 18 = 0

2(x + 3)2 = 18

Dividing by 2 on both sides.

(2(x + 3)2)/2=18/2

(x + 3)2 = 9

Using square root property.

x + 3 = √9

x + 3 = ±3

x + 3 = 3

x = 3 - 3

x = 0

x + 3 = -3

x = -3 - 3

x = -6

So, the solution is {-6, 0}.

Problem 8 :

(x - 6)2 – 4 = 14

Solution :

(x - 6)2 – 4 = 14

(x - 6)2 = 14 + 4

(x - 6)2 = 18

Using square root property

x - 6 = √18

x - 6 = √(9 × 2)

x – 6 =√( 3  2)

x – 6 = ±3√2

x – 6 = 3√2

x = 3√2 + 6

x – 6 = -3√2

x = 3√2 - 6

So, the solution is {3√2 + 6, 3√2 - 6}.

Problem 9 :

5(x + 6)2 + 60 = 0

Solution :

5(x + 6)2 + 60 = 0

5(x + 6)2 = -60

Dividing by 5 on both sides.

(5(x + 6)2)/5 = -60/5

(x + 6)2 = -12

Using square root property.

 x + 6 = √-12

So, there is no real root.

Problem 10 :

4(x + 4)2 = 9

Solution :

4(x + 4)2 = 9

Dividing by 4 on both sides.

4(x + 4)2)/4 = 9/4

(x + 4)2 = 2.25

Using square root property.

 x + 4 = √2.25

x + 4 = ±1.5

x + 4 = 1.5

x = 1.5 - 4

x = -2.5

x + 4 = -1.5

x = -1.5 - 4

x = -5.5

So, the solution is {-2.5, -5.5}.

Problem 11 :

4(x - 2)2 + 15 = 0

Solution :

4(x - 2)2 + 15 = 0

4(x - 2)2 = -15

Dividing by 4 on both sides.

(4(x - 2)2)/4 = -15/4

(x - 2)2 = -15/4

Using square root property.

 x - 2 = √(-15/4)

So, there is no real root.

Problem 12 :

5(x - 2)2 – 45 = 0

Solution :

5(x – 2)2 – 45 = 0

5(x - 2)2 = 45

Dividing by 5 on both sides.

(5(x - 2)2)/5 = 45/5

(x - 2)2 = 9

Using square root property.

 x - 2 = √9

x - 2 = ±3

x - 2 = 3

x = 3 + 2

x = 5

x - 2 = -3

x = -3 + 2

x = -1

So, the solution is {-1, 5}.

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