Word Problems on Quadratic Equations :
In this section, we are going to learn, how to solve quadratic equation word problems step by step.
Problem 1 :
Difference between a number and its positive square root is 12. Find the number.
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?
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
[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.
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
(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.
Let "x" and "x + 4" be the lengths of other two sides.
Using Pythagorean theorem, we have
(x + 4)² + x² = 20²
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.
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 side of the equilateral triangle is 17 units.
Apart from the problems given above, if you need more word problems on quadratic equations
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