**Radical Equations**

**Unit 7 Problems**

**Application problems including radical equations.**

1. A positive real number is 4 less than another. When 8 times the larger is added to the square of the smaller, the result is 96. Find the numbers.

__Solution__

Let one number be x

The other number is x+4

8(x+4) + x^{2} = 96

8x+32x^{2} = 96

x^{2} + 8x - 64= 0

x^{2} + 8x +16 = 80

(x+4)^{2} = 80

x = -4-4√5

or x = -4+4√5

2. A positive real number is 4 less than another. If the sum of the squares of the two numbers is 72, then find the numbers.

__Solution__

Let one number be x

The other number is x+4

(x+4)^{2} + x^{2} = 72

x^{2}+x^{2} + 8x + 16 = 72

x^{2} + 4x – 28 = 0

x^{2} + 4x = 28

(x+2)^{2} = 32

x = -2±√32

x= -2+√32

or x = -2+√32

3. The area of a rectangle is 46 square inches. If the length is 4 times the width, then find the dimensions of the rectangle. Round off your answers to the nearest hundredth.

__Solution__

x(4x) = 46

x^{2} - 46 = 0

4x^{2} = 46

x^{2} = 23

x^{2} = 11.5

x = -3.39

or

x = 3.39

ignore -3.39

Hence, the dimensions are;

Length = 3.39x4 =13.56 Inches

Width = 3.39 Inches.

4. If the sides of a square measure units, then find the length of the diagonal.

__Solution__

Let the length of the diagonal be x

X^{2} = (9√3)^{2} + (9√3)^{2}

X^{2} = (81x3) + (81x3)

X^{2} = 486

X = 22.05 Units

5. The area of a rectangle is 20 square inches. If the length is 4 inches less than 6 times the width, then find the dimensions of the rectangle. Round off your answers to the nearest hundredth.

__Solution__

Let the width be x

Length = 6x-4

x(6x-4) = 20

x^{2}-x-=0

(x-)=

Either x = -1.52(ignore)

Or x = 2.19

Length = 6(2.19)-4 = 9.14 Inches.

Width = 2.19 Inches.

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