WORD PROBLEMS WITH QUADRATIC EQUATIONS EXAMPLES

Word Problems with Quadratic Equations Examples :

In this section, let us see some examples on solving word problems with quadratic equations.

Solving Word Problems with Quadratic Equations Examples

Example 1 :

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution :

Let x be the length of shorter side of the rectangle

Length of diagonal  =  x + 60

Length of longer side  =  x + 30

(x + 60)²   =  x² + (x + 30)²

x² + 60² + 2 x (60)  =  x² + x² + 2 x (30) + 30²

x² + 3600 + 120 x  =  2 x² + 60 x + 900

2 x² - x² + 60 x - 120 x + 900 - 3600  =  0

x² - 60 x - 2700  =  0

x² - 90 x + 30 x - 2700  =  0

x (x - 90) + 30 (x - 90)  =  0

 (x + 30) (x - 90)  =  0

  x + 30  =  0           x - 90  =  0

   x  =  -30                x  =  90

Breadth of rectangle  =  90 m

Length of rectangle  =  90 + 30  =  120 m

Example 2 :

The  difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution :

Let 'x' be the larger number

let 'y' be the smaller number

y²  =  8 x

The difference of squares of two numbers = 180

 x² - y²  =  180

  x² - 8 x  =  180

  x² - 8 x - 180  =  0

  x² - 18 x + 10 x - 180  =  0

x (x - 18) + 10 (x - 18)  =  0

 (x - 18) (x + 10)  =  0 

 x - 18  =  0         x + 10  =  0

 x  =  18             x  =  -10

y²  =  8 (18)

 y  =  8 (18)

 y  =  √2 · · 2 · 3 · 3 · 2

 y  =  2 · 2 · 3

 y  =  12

Hence the larger number  =  18 and smaller number = 12.

Example 2 :

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more,it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution :

Let x be the speed of the train

If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey

Time  =  distance/speed

Distance to be covered  =  360 km

 T₁  =  360/x

 T₂  =  360/(x + 5)

 T₁ - T₂  =  1 hour

 [360/x] -  [360/(x + 5)]  =  1

 360 [ (1/x) - (1/(x + 5))]  =  1

 360 [ (x + 5 - x)/x(x + 5) ]  =  1

 360[5/(x² + 5 x)]  =  1

1800/(x² + 5 x)  =  1

1800  =  (x² + 5 x)

x² + 5 x - 1800  =  0

x² - 40 x + 45 x - 1800  =  0

x (x - 40) + 45 (x - 40)  =  0

(x - 40) (x + 45)  =  0

 x - 40  =  0     x + 45  =  0

 x  =  40   and x  =  -45

Therefore speed of the train  =  40 km/hr

After having gone through the stuff given above, we hope that the students would have understood, word problems with quadratic equations examples.

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