SOLVING QUADRATIC EQUATIONS USING SQUARE ROOTS

Solve the following quadratic equations using square root :

Example 1 :

x² - 2 = 23

Solution :

x² - 2 = 23

Add 2 to each side.

x² = 25

Take square root on both sides.

√x² = ±√25

x = ±5

Therefore, the solutions are

x = 5 and x = -5

Example 2 :

(x - 2)² = 36

Solution :

(x - 2)² = 36

Take square root on both sides.

√(x - 2)² = ±√36

x - 2 = ±6

Therefore, the solutions are 

x = 8 and x = -4

Example 3 :

3x² + 8 = 56

Solution :

3x² + 8 = 56

Subtract 8 from both sides.

3x² = 48

Divide both sides by 3.

x² = 16

Take square root on both sides.

√x² = ±√16

x = ±4

Therefore, the solutions are

x = 4 and x = -4

Example 4 :

-5x² + 15 = -2x² - 12

Solution :

-5x² + 15 = -2x² - 12

Add 2x² to both sides.

-3x² + 15 = -12

Subtract 15 from both sides.

-3x² = -27

Divide both sides by -3.

x² = 9

Take square root on both sides.

√x² = ±√9

x = ±3

Therefore, the solutions are

x = 3 and x = -3

Example 5 :

6(x² - 1) + 7(1 - x²) = -11

Solution :

6(x² - 1) + 7(1 - x²) = -11

Use Distributive Property.

6x² - 6 + 7 - 7x² = -11

Group the like terms.

(6x² - 7x²) + (-6 + 7) = -11

Combine the like terms.

-x² + 1 = -11

Subtract 1 from both sides.

-x² = -12

Multiply both sides by -1.

x² = 12

Take square root on both sides.

√x² = ±√12

x = ±√(2 ⋅ 2 ⋅ 3)

x = ±2√3

Therefore, the solutions are

x = 2√3 and x = -2√3

Example 6 :

-6(x² - 10)² - 5 = -221

Solution :

-6(x² - 10)² - 5 = -221

Add 6 to both sides.

-6(x² - 10)² = -216

Divide both sides by -6.

(x² - 10)² = 36

Take square root on both sides.

√(x² - 10)² = ±√36

x² - 10 = ±6

Therefore, the solutions are

x = 4, x = -4, x = 2 and x = -2

Example 7 :

x² + 6x + 8 = 0

Solution :

x² + 6x + 8 = 0

Write the trinomial on the left side of the above equation in terms of square of a binomial.

x² + 2(x)(3) + 8 = 0

x² + 2(x)(3) + 3² - 3² + 8 = 0

Using the identity (a + b)² = a² + 2ab + b²,

(x + 3)² - 3² + 8 = 0

(x + 3)² - 9 + 8 = 0

(x + 3)² - 1 = 0

Add 1 to both sides.

(x + 3)² = 1

Take square root on both sides.

√(x + 3)² = ±√1

x + 3 = ±1

Therefore, the solutions are

x = -2 and x = -4

Example 8 :

x² - 5x + 6 = 0

Solution :

x² - 5x + 6 = 0

Write the trinomial on the left side of the above equation in terms of square of a binomial.

x² - 2(x)(5/2) + 6 = 0

x² - 2(x)(5/2) + (5/2)² - (5/2)² + 6 = 0

Using the identity (a - b)² = a² - 2ab + b²,

(x - 5/2)² - (5/2)² + 6 = 0

(x - 5/2)² - 25/4 + 6 = 0

(x - 5/2)² - 1/4 = 0

Add 1/4 to both sides.

(x - 5/2)² = 1/4

Take square root on both sides.

√(x - 5/2)² = ±√1/4

x - 5/2 = ±1/2

Therefore, the solutions are

x = 3 and x = 2

Example 9 :

A touch tank has a height of 3 feet. Its length is three times its width. The volume of the tank is 270 cubic feet. Find the length and width of the tank.

Solution :

Volume of the tank = 270 cubic feet

height = 3 feet

Let w be the width.

length = 3w

length x width x height = 270

3(w)(3w) = 270

9w² = 270

w² = 270/9

w = ±√30

Example 10 :

The area A of an equilateral triangle with side length s is given by the formula A = (√3/4)  s². Solve the formula for s. Then approximate the side length of the traffic sign that has an area of 390 square inches.

A = (√3/4)  s²

Area of equilateral triangle = 390 square inches

390 = (√3/4) s²

390(4/√3) = s²

s² = 390(4/√3) x (√3/√3)

= 390 x 4 x (√3/3)

= 130 x 4 √3

s² = 520√3

The side length of the traffic sign is about 30 inches.

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