# SQUARE ROOT PROPERTY

Taking the square root on both sides of an equation results in a positive and a negative answer. This is called the Square Root Property.

Let y be a variable and d be a positive real number.

Then the equation y2 = d has exactly two solutions :

y = ±√d

When we take sqaure root on both sides of an equation in the form x2 = k, why does it result in a positive and a negative answer?

Consider the following equation.

x2 = 4

Is the above equation true only for x = 2?

No, the above equation is true for two values of x.

x = -2 or x = 2

(-2)2 = 4

22 = 4

This is the reason, we consider both positive and negative signs for square root of a number.

Example 1 :

Solve for x :

x2 = 9

Solution :

x2 = 9

Take square root on both sides.

x2 = ±√9

x = ±√(3 ⋅ 3)

x = ±3

x = -3  or  x = 3

Example 2 :

Solve for k :

k2 + 5 = 5

Solution :

k2 + 5 = 5

Subtract 5 from both sides.

k2 = 0

Take square root on both sides.

√k2 = ±√0

k = 0

Example 3 :

Solve for y :

3y2 = 147

Solution :

3y2 = 147

Divide both sides by 3.

y2 = 49

Take square root on both sides.

y2 = ±√49

y = ±√(7 ⋅ 7)

y = ±7

y = -7  or  y = 7

Example 4 :

Solve for v :

25v2 = 1

Solution :

25v2 = 1

Divide both sides by 25.

v2 = ¹⁄₂₅

Take square root on both sides.

√v2 = ±√¹⁄₂₅

v = ±√(¹⁄₅ ⋅ ¹⁄₅)

v = ±¹⁄₅

v = -¹⁄₅  or  v = ¹⁄₅

Example 5 :

Solve for n :

n2 + 4 = 40

Solution :

n2 + 4 = 40

Subtract 4 from both sides.

n2 = 36

Take square root on both sides.

√n2 = ±√36

n = ±√(⋅ 6)

n = -6  or  n = 6

Example 6 :

Solve for x :

x2 - 2 = 17

Solution :

x2 - 2 = 17

x2 = 19

Take square root on both sides.

√x2 = ±√19

x = ±√19

x = -√19  or  x = √19

Example 7 :

Solve for b :

4b2 + 2 = 326

Solution :

4b2 + 2 = 326

Subtract 2 from both sides.

4b= 324

Divide both sides by 4.

b= 81

Take square root on both sides.

√b2 = ±√81

b = ±√(9 ⋅ 9)

b = ±9

b = -9  or  b = 9

Example 8 :

Solve for z :

(2z2 + 7)/3 = 19

Solution :

(2z2 + 7)/3 = 19

Multiply both sides by 3.

2z2 + 7 = 57

Subtract 7 from both sides.

2z2 = 50

Divide both sides by 2.

z2 = 25

Take square root on both sides.

√z2 = ±√25

z = ±√(5 ⋅ 5)

z = ±5

z = -5  or  z = 5

Example 9 :

Solve for b :

(b + )2 -  = 0

Solution :

(b + )2 -  = 0

(b + )2 =

Take square root on both sides.

(b + )2 = ±√

b +  = ±√(⋅ )

b +  = ±

b +  = ⁻¹⁄₃  or  b +  =

 b + ⅓ = ⁻¹⁄₃b = ⁻¹⁄₃ ⁻ ¹⁄₃b = ⁻²⁄₃ b + ⅓ = ¹⁄₃b = ¹⁄₃ ⁻ ¹⁄₃b = 0

Example 10 :

Solve for r :

9r2 - 5 = 607

Solution :

9r2 - 5 = 607

9r2 = 612

Divide both sides by 9.

r2 = 68

Take square root on both sides.

√r2 = ±√68

r = ±√(2 ⋅ 2 ⋅ 17)

r = ±2√7

r = -2√17  or  r = 2√17

Example 11 :

Solve for x :

x2 + 9 = 0

Solution :

x2 + 9 = 0

Subtract 9 from both sides.

x2 = -9

Take square root on both sides.

√x2 = ±√-9

x = ±√(-1 ⋅ 9)

x = ±√(i2 ⋅ 3 ⋅ 3)

x = ±3i

x = -3i  or  x = 3i

Example 12 :

Solve for y :

10 - 5y2 = -310

Solution :

10 - 5y2 = -310

Subtract 10 from both sides.

-5y2 = -320

Divide both sides by -5.

y2 = 64

Take square root on both sides.

√y2 = ±√64

y = ±√(8 ⋅ 8)

y = ±8

y = -8  or  y = 8

Example 13 :

Solve for z :

5 - 3z2 = 41

Solution :

5 - 3z2 = 41

Subtract 5 from both sides.

-3z2 = 36

Divide both sides by -3.

z2 = -36

Take square root on both sides.

√z2 = ±√-36

z = ±√(-1 ⋅ 36)

z = ±√(-1 ⋅ 6 ⋅ 6)

z = ±√(i2 ⋅ 6 ⋅ 6)

z = ±6i

z = -6i  or  z = 6i

Example 14 :

Solve for x :

x2 + 6x + 9 = 0

Solution :

x2 + 6x + 9 = 0

In the expression on the left side of the equation above, there are two variable terms, one of which contains square.

To use square root property, we have to rewrite the expression such that it has to contain only one variable term.

Rewrite x2 + 6x + 9 in the form of a2 + 2ab + b2.

x2 + 2(x)(3) + 32 = 0

We can use the algebraic identity (a + b)2 a2 + 2ab + b2  to write the expression on the left side in terms of square of a binomial.

(x + 3)2 = 0

Take square root on both sides.

(x + 3)2 = ±√0

x + 3 = 0

x = -3

Example 15 :

Solve for y :

y2 - 16y + 15 = 0

Solution :

y2 - 16y + 15 = 0

Since there are two variable terms in the expression on the left side, we have to solve the equation as done in example 14 above.

Rewrite y2 - 16y + 15 in the form of a2 - 2ab + b2.

y2 - 2(y)(8) + 15 = 0

y2 - 2(y)(8) + 8- 8+ 15 = 0

y2 - 2(y)(8) + 8- 64 + 15 = 0

y2 - 2(y)(8) + 8- 49 = 0

We can use the algebraic identity (a - b)2 a2 - 2ab + b2  to write the expression on the left side in terms of square of a binomial.

(y - 8)2 - 49 = 0

(y - 8)2 = 49

Take square root on both sides.

(y - 8)2 = ±√49

y - 8 = ±√(7 ⋅ 7)

y - 8 = ±7

 y - 8 = -7y = 1 y - 8 = 7y = 15

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