**Quadratic Equation Word Problems Worksheet with Answers :**

Worksheet given in this section is much useful to the students who would like to practice solving word problems on quadratic equation.

**Problem 1 :**

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

**Problem 2 :**

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 3 :**

Divide 25 in two parts so that sum of their reciprocals is 1/6.

**Problem 4 :**

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 5 : **

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.

**Problem 1 :**

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

**Solution : **

Let "x" be the required number.

Its positive square root is √x

**Given :** Difference between x and √x = 12

x - √x = 12

x - 12 = √x

(x - 12)² = x

x² - 24x + 144 = x

x² - 25x + 144 = 0

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

x = 9 or x = 16

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

Hence the required number is 16.

**Problem 2 :**

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 ?

**Solution :**

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

Simplify.

[60x - 60(x - 2)] / [x(x - 2)] = 1

[60x - 60x + 120] / [x² - 2x] = 1

120 / (x² - 2x) = 1

120 = x² - 2x

0 = x² + 2x - 120

x² + 2x - 120 = 0

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

x = - 10 or x = 12

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

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

**Problem 3 :**

Divide 25 in two parts so that sum of their reciprocals is 1/6.

**Solution :**

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.

Then, we have

1/x + 1/(25 - x) = 1/6

Simplify.

(25 - x + x) / x(25 - x) = 1/6

25 / (25x - x²) = 1/6

6(25) = 25x - x²

150 = 25x - x²

x² - 25x + 150 = 0

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

x = 15 or x = 10

When x = 15,

25 - x = 25 - 15

25 - x = 10

When x = 10,

25 - x = 25 - 10

25 - x = 15

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

**Problem 4 :**

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.

**Solution :**

Let "x" and "x + 4" be the lengths of other two sides.

Using Pythagorean theorem, we have

(x + 4)² + x² = 20²

Simplify.

x² + 8x + 16 + x² = 400

2x² + 8x + 16 = 400

Subtract 400 from both sides.

2x² + 8x - 384 = 0

Divide both sides by 2.

x² + 4x - 192 = 0

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

x = -16 or x = 12

x = -16 can not be accepted. Because length can not be negative.

If x = 12,

x + 4 = 12 + 4 = 16

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

**Problem 5 : **

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.

**Solution :**

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

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).

Using Pythagorean theorem, we have

(x - 12)² = (x - 13)² + (x - 14)²

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

x² - 30x + 221 = 0

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

x = 13 or x = 17.

x = 13 can not be accepted.

Because, if x = 13, the side represented by (x - 14) will be negative.

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

Apart from the problems given above, if you need more word problems on quadratic equations

After having gone through the stuff given above, we hope that the students would have understood "Quadratic equation word problems worksheet with answers"

Apart from the stuff given above, if you want to know more about "Quadratic equation word problems worksheet with answers", please click here.

Apart from the stuff given on "Quadratic equation word problems worksheet with answers", if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

**WORD PROBLEMS**

**HCF and LCM word problems**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**

**Sum of all three four digit numbers formed using 0, 1, 2, 3**

**Sum of all three four digit numbers formed using 1, 2, 5, 6**