Solve each problem over R Approximate any answers involving

Solve each problem over R. Approximate any answers involving radicals to the nearest tenth. A rug is to cover two-thirds of the floor area of a room 20 feet by 30 feet. The uncovered part of the floor is to form a strip of uniform width around the rug. Find the width of the strip. The volume of a rectangular jewelry box 4 cm in height is 136 square centimeters. If the perimeter of the base is 24 centimeters, find the dimensions of the base. A rectangular corner lot has dimensions 20 meters. When two adjoining streets are each widened by the same amount, a quarter of the area is last. Find the dimensions of the lot. The base of a triangle is 6 meters greater than its altitude and the area of the triangle is 4 square meters. Find the length of the base. A vegetable garden measures 9 feet by 12 feet. By what equal amount must each dimension be increased if the area is to be doubled? One leg of a right triangle is 3 cm longer than the other. The hypotenuse is 8 centimeters long. Find the length of the legs and the area of the triangle. A second number is 2 less than a first number. Their product is 4. Find the numbers. One number is 2 less than twice another number. Find these numbers if their product is 3. Two numbers differ by 2. Their reciprocals add to 3. Find the numbers. Solve over C. Use the most appropriate most appropriate method. y^2 + 4 squareroot 2y + 16 = 0 x - 1/2 - 5/2 = 2/1 - x (x + 1/x + 2)^2 + 2(x + 1/x + 2) = 8 4x^2 = 11 + 4x x^-2 - 5x^-1 -6 = 0 x^2 + 6ix - 8 = 0 3x - 2 squareroot x = 8 Find a quadratic equation having the given roots. {3 plusorminus 2 squareroot 5} {-1/2, 2/3} {2 plusorminus squareroot 2/3}

Solution

1) Let the width of strip be x

rug covers 2/3 rd area = (2/3)(20*30) = 2*20*10 = 400 sq ft

Length of rug after width of strip is deducted = 20 -2x

Width of rug after width of strip is deducted = 30 -2x

Area = (20 -2x)( 30 -2x) = 400

2*2(10 - x)(15 -x) = 400

(10 -x)(15 -x) = 100

x^2 - 25x + 150 = 100

x^2 - 25x + 50 =0

Solve the quadratic using quadratic formula : ax^2 + bx + c=0

x = ( - b +/- sqrt( b^2 - 4ac) )/2a

we get x= 22.80 ; x= 2.19 .Now x is the width of strip and it cannot be greater than dimesnions of floor

So, correct solution is width of strip = 2.19 feet

Use quadratic formula:

1) y^2 + 4sqrt2y + 16 =0

Solve the quadratic using quadratic formula : ax^2 + bx + c=0

x = ( - b +/- sqrt( b^2 - 4ac) )/2a

y = ( - 4sqrt2 +/- sqrt( 32 -4*16) )/2

= -2sqrt2 + /- 2isqrt2