QUADRATIC EQUATION WORD PROBLEMS WORKSHEET WITH ANSWERS

Problem 1 :

Difference between a number and its positive square root is 12. Find the number.

Problem 2 :

If the difference between a number and its reciprocal is ²⁴⁄₅, find the number.

Problem 3 :

A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod?

Problem 4 :

Divide 25 in two parts so that sum of their reciprocals is .

Problem 5 :

The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Problem 6 :

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Problem 7 :

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article(in dollars) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was $90, find the number of articles produced and the cost of each article.

Problem 8 : 

The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. Find the length of each side of the equilateral triangle.

Answers

1. Answer :

Let x be the required number.

Its positive square root is √x. 

Given : Difference between x and √x is 12.

x - √x = 12

x - 12 = √x

(x - 12)2 = x

x2 - 24x + 144 = x

Subtract x from both sides.

x2 - 25x + 144 = 0

x2 - 9x - 16x + 144 = 0

x(x - 9) - 16(x - 9) = 0

(x - 9)(x - 16) = 0

x = 9  or  x = 16

9 - √9 = 9 - 3

= 6 ≠ 12

16 - √16 = 16 - 4

= 12

x = 9 does not satisfy the condition given in the question.

Then, 

x  =  16

Therefore, the required number is 16.

2. Answer :

Let y be the required number.

Then, its reciprocal is ¹⁄y.

Given : Difference between a number and its reciprocal is ²⁴⁄₅.

y - ¹⁄y ²⁴⁄₅

Multiply both sides by 5y to get rid of the denominators y and 5.

5y(y - ¹⁄y) = 5y(²⁴⁄₅)

5y(y) - 5y(¹⁄y) = 24y 

5y2 - 5 = 24y

Subtract 24y from boths sides.

5y2 - 24y - 5 = 0

Solve by factoring.

5y2 - 25y + y - 5 = 0

5y(y - 5) + 1(y - 5) = 0

(y - 5)(5y + 1) = 0

y - 5 = 0  or  5y + 1 = 0

y = 5  or  y = ⁻¹⁄₅

Justification :

5 - ⅕ ²⁵⁄₅ -

⁽²⁵ ⁻ ¹⁾⁄₅

²⁴⁄₅

⁻¹⁄₅ - (-5) ⁻¹⁄₅ + 5 

⁻¹⁄₅ + ²⁵⁄₅

⁽⁻¹ ⁺ ²⁵⁾⁄₅

²⁴⁄₅

Both the values y = 5  and y = ⁻¹⁄₅ satisfy the condition given in the question.

Therefore, the required number is 5 or ⁻¹⁄₅.

3. Answer :

Let x be the length of the given rod.

Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). 

Cost of one meter of the given rod  is

⁶⁰⁄ₓ 

Cost of one meter of the rod which is 2 meter shorter is

⁶⁰⁄₍ₓ ₋ ₂₎

Given : If the rod was 2 meter shorter and each meter costs $1 more.

That is, 60/(x-2) is $1 more than 60/x.

⁶⁰⁄₍ₓ ₋ ₂₎ - ⁶⁰⁄ₓ = 1

Multiply both sides by x(x - 2) to get rid of the denominators (x - 2) an x.

x(x - 2)[⁶⁰⁄₍ₓ ₋ ₂₎ - ⁶⁰⁄ₓ] = x(x - 2)

x(x - 2)[⁶⁰⁄₍ₓ ₋ ₂₎] - x(x - 2)[⁶⁰⁄ₓ] = x(x - 2)

60x - 60(x - 2) = x- 2x

60x - 60x + 120 = x- 2x

120 = x- 2x

0 = x2 - 2x - 120 

x2 - 2x - 120 = 0

Solve by factoring.

x2 - 12x + 10x - 120 = 0

x(x - 12) + 10(x - 12) = 0

(x - 12)(x + 10) = 0

x - 12 = 0  or  x + 10 = 0

x = 12  or  x = -10

Because length can not be a negative number, we can ignore x = -10. 

Therefore, the length of the given rod is 12 m.

4. Answer :

Let x be one of the parts of 25.

Then, the other part is (25 - x). 

Given : Sum of the reciprocals of the parts is 1/6. 

¹⁄ₓ + ¹⁄₍₂₅ ₋ ₓ₎

Multiply both sides by 6x(25 - x) to get rid of the denominators x, (25 - x) and 6.

6x(25 - x)[¹⁄ₓ + ¹⁄₍₂₅ ₋ ₓ₎] = 6x(25 - x)[⅙]

6x(25 - x)[¹⁄ₓ] - 6x(25 - x)[¹⁄₍₂₅ ₋ ₓ₎] = x(25 - x)

6(25 - x) - 6x = 25x - x2

150 - 6x + 6x = 25x - x2

150 = 25x - x2

x2 - 25x + 150 = 0

Solve by factoring.

x2 - 15x - 10x + 150 = 0

x(x - 15) - 10(x - 15) = 0

(x - 15)(x - 10) = 0

x - 15 = 0  or x - 10 = 0

x = 15  or  x = 10

When x = 15, 

25 - x = 25 - 15

= 10

When x = 10,

25 - x = 25 - 10

= 15

Therefore, the two parts of 25 are 10 and 15.

5. Answer :

From the given information. we can assume x and (x + 4) as the lengths of the other two sides of the right triangle.

triangle5

Using Pythagorean Theorem in the right triangle above,

x+ (x + 4)2 = 202

x+ (x + 4)(x + 4) = 400

x+ x2 + 4x + 4x + 16 = 400

2x2 + 8x + 16 = 400

2x2 + 8x - 384 = 0

Divide both sides by 2. 

x2 + 4x - 192 = 0

Solve by factoring.

x2 - 12x + 16x - 192 = 0

x(x - 12) + 16(x - 12) = 0

(x - 12)(x + 16) = 0

x - 12 = 0  or  x + 16 = 0

x - 12 = 0  or  x + 16 = 0

x = 12  or  x = -16

x = -16 can not be accepted. Because length can never be negative.

So,

x = 12

x + 4 = 16

Therefore, the other two sides of the triangle are 12 cm and 16 cm.

6. Answer :

Let x be the base the right triangle.

Then, its altitude is (x - 7).

triangle4

Using Pythagorean Theorem in the right triangle above,

x2 + (x - 7)2 = 132

x2 + (x - 7)(x - 7) = 169

x2 + x2 - 7x - 7x + 49 = 169

2x2 - 14x + 49 = 169

2x- 14x - 120 = 0

Divide both sides by 2.

x- 7x - 60 = 0

Solve by factoring.

x- 12x + 5x - 60 = 0

x(x - 12) + 5(x - 12) = 0

(x + 5)(x - 12) = 0

x + 5 = 0  or  x - 12 = 0

x = -5  or  x = 12

Because x represents the base of the triangle, it can never be negative. So, x = 12 

Base = 12 cm

 Altitude = 5 cm

Therefore, the other two sides of the right triangle are 12 cm and 5 cm.

7. Answer :

Let x be the number of articles produced on that day.

Then, the cost of production of each article is

= 2x + 3

Total cost = Number of articles  cost of one article

90 = x(2x + 3)

90 = 2x2 + 3x

2x2 + 3x - 90 = 0

Solv by factoring.

2x2 - 12x + 15 x - 90 = 0

2x(x - 6) + 15(x - 6) = 0

(2x + 15)(x - 6) = 0

2x + 15 = 0  or  x - 6 = 0

x = -7.5  or  x = 6

Because x represents the number of articles, it can never be negative value. Then, x = 6.

So, the number of articles produced is 6.

The cost of each article : 

= 2x + 3

= 2(6) + 3

= 12 + 3

= $15

8. Answer :

Let x be the length of each side of the equilateral triangle.

triangle6

Then, the sides of the right angle triangle are

(x - 12), (x - 13) and (x - 14) 

In the above three sides, the side represented by (x - 12) is hypotenuse (because, that is the longest side).

triangle7

Using Pythagorean Theorem in the right triangle above,

(x - 13)2 + (x - 14)= (x - 12)2

(x - 13)(x - 13) + (x - 14)(x - 14) = (x - 12)(x - 12)

x2 - 26x + 169 + x2 - 28x + 196 = x2 - 24x + 144

2x2 - 54x + 365 = x2 - 24x + 144

x2 - 30x + 221 = 0

x2 - 30x + 221 = 0

Solve by factoring.

x2 - 13x - 17x + 221 = 0

x(x - 13) - 17(x - 13) = 0

(x - 13)(x - 17) = 0

x - 13 = 0  or  x - 17 = 0

x = 13  or  x = 17 

x = 13 can not be accepted.

Because, if x = 13, the side represented by (x - 14) is negative. So, x = 17.

Therefore, the length of each side of the equilateral triangle is 17 units.

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