Word Problems on Quadratic Equations with Answers :
In this section, we will see how to solve word problems involving the concept quadratic equation.
Example 1 :
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let 'x' be the base of the triangle
Altitude = x - 7
(Hypotenuse side)² = (Base)² + (Height)²
13² = x² + (x - 7)²
169 = x² + x² - 2 (x) (7) + 7²
169 = x² + x² - 14 x + 49
0 = 2 x²- 14 x + 49 - 169
2 x²- 14 x - 120 = 0
dividing the whole equation by 2 => x²- 7 x - 60 = 0
x²- 12 x + 5 x - 60 = 0
x (x - 12) + 5 (x - 12) = 0
(x + 5) (x - 12) = 0
x + 5 = 0 x - 12 = 0
x = -5 x = 12
Base = 12 cm
Height = 12 - 7 = 5 cm
Example 2 :
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 rupees) 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.
Let 'x' be the number of articles produced on that day
cost of production of each article = 2 x + 3
Total cost of production = number of article x cost of one article
90 = x (2 x + 3)
90 = 2 x² + 3 x
2x² + 3 x - 90 = 0
2x² - 12 x + 15 x - 90 = 0
2x (x - 6) + 15 (x - 6) = 0
(2 x + 15) (x - 6) = 0
2x + 15 = 0 x - 6 = 0
2x = -15 x = 6
x = -15/2
Hence the number of articles produced on that particular day = 6.
After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems on quadratic equations.
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