Quadratic equation word problems area much useful to the students who would like to practice problems on quadratic equations.

Let us some quadratic equation word problems.

**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.**

Let us look at the next word problem on "Quadratic equation word problems"

**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 = 60/x

Cost of one meter of the rod which is 2 meter shorter = 60/(x-2)

From the question, 60/(x-2) = (60/x) + 1

60/(x-2) = (60+x)/x

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

60x = x² + 58x - 120

x² -2x -120 = 0

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

x = -10 or x = 12

x = -10 can not be accepted. Because length can not be negative.

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

**Let us look at the next word problem on "Quadratic equation word problems"**

**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).

Sum of the reciprocals of the parts = 1/6

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

(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

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

**Let us look at the next word problem on "Quadratic equation word problems"**

**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, (x+4)² + x² = 202

x² + 8x + 16 + x² - 400 = 0

2x² + 8x - 384 = 0

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.**

**Let us look at the next word problem on "Quadratic equation word problems"**

**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. The side of the equilateral triangle is

**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, (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, one of the sides of the right angle triangle would be negative. **Hence, the side of the equilateral triangle is 17 units.**

**Let us look at the next word problem on "Quadratic equation word problems"**

**Problem 6 :**

The area of a rectangular field is 2000 sq.m and its perimeter is 180 m. Find the length and width of the field.

**Solution :**

Let "x" and "y" be the length and width of the rectangle respectively.

Perimeter = 180 m (given)

2x + 2y = 180 ---> x + y = 90

So, y = 90 - x

Area = 2000 sq.m

xy = 2000

x(90-x) = 2000

90x - x² = 2000

x² - 90x + 2000 = 0

(x-50)(x-40) = 0

x = 50 or x = 40

If x = 50, y = 90 - 50 = 40

If x = 40, y = 90 - 40 = 50 **Hence, the length and breadth of the rectangle are 40 m and 50 m respectively.**

**Let us look at the next word problem on "Quadratic equation word problems"**

**Problem 7 :**

A distributor of apple juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D is given by D = -2000p² + 2000p + 17000. Find the price per bottle that will result zero inventory.

**Solution :**

Stock in the store is 5000 bottles. If the inventory be zero, the demand has to be 5000 bottles.

To know the price for zero inventory, we have to plug D = 5000 in the quadratic equation D = -2000p2 + 2000p + 17000

5000 = -2000p² + 2000p + 17000

2000p² - 2000p - 12000 = 0

p² - p - 6 = 0

(p + 3)(p - 2) = 0

p = -3 or p = 2

p = -3 can not be accepted. Because, price can not be negative. **Hence, price per bottle that will result zero inventory is $2.**

**Let us look at the next word problem on "Quadratic equation word problems"**

**Problem 8 :**

Two squares have sides p cm and (p+5) cm. The sum of their squares is 625 sq.cm. Find the sides of the squares.

**Solution :**

**According to the question, we have**

**p² + (**p+5)² = 625

**p² + **p² + 10p + 25 = 625

**2p² + **10p - 600 = 0

**p² + **5p - 300 = 0

(p -15)(p + 20) = 0

p = 15 or p = -20

p = -20 can not be accepted. Because, the side of the square can not be negative.

If p = 15, p + 5 = 15 + 5 = 20

**Hence, the sides of the square are 15 cm and 20 cm.**

**Let us look at the next word problem on "Quadratic equation word problems"**

**Problem 9 :**

A ball is thrown upwards from a rooftop which is above from the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball "h" from the ground at time "t" seconds is given by, h = -16t**²** + 64t + 80.How long will the ball take to hit the ground?

**Solution :**

**When the ball hits the ground, height "h" = 0.**

So, we have

0 = -16t**²** + 64t + 80

16t**²** - 64t - 80 = 0

t**²** - 4t - 5 = 0

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

t = 5 or t = -1

t = -1 can not be accepted. Because time can never be a negative value.

**Hence, the ball will take 5 seconds to hit the ground.**

**Let us look at the next word problem on "Quadratic equation word problems"**

**Problem 10 :**

A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

**Solution :**

**After enlargement, let "x" be the width of picture. **

**Then, the height of the picture = 4x / 3**

**(From the given information, it is very clear that the picture is in the shape of rectangle. So, the given height can be considered as length of the rectangle) **

**Given : Area after enlargement = 192 square inches. **

**So, we have **

**Length X width = 192 sq. inches**

(4x/3)(x) = 192

4x**²**/3 = 192

x**²** = 192(3/4)

x**²** = 144

x = 12

**Therefore the width = 12 inches. **

**The height = 4(12) / 3 = 16 inches**

**Hence, the dimensions of the picture are 12 inches and 16 inches. **

After having gone through the step by step solutions for all the problems on "Quadratic equation word problems", we hope that the students would have understood how to solve quadratic equation word problems.

If you want to know more about "Quadratic equation word problems", please click 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**