WORD PROBLEMS ON QUADRATIC EQUATIONS WITH ANSWERS

Word Problems on Quadratic Equations with Answers :

In this section, we will see how to solve word problems involving the concept quadratic equation.

Word Problems on Quadratic Equations with Answers

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.

Solution :

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.

Solution :

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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