## WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS BY FACTORING

Problem 1 :

Two natural numbers differ by 2 and their product is 360. Find the numbers.

Solution :

Let x and y be two natural numbers

it differs by 2

So,

x - y  =  2 -----(1)

Their product is 360

So,

xy  =  360

y  =  360/x ----- (2)

Now we may apply the value of y in the first equation.

x - (360/x)  =  2

(x2 - 360)/x  =  2

x2 - 360  =  2x

x2 - 2x - 360  =  0

(x + 18)(x - 20)  =  0

 x + 18  =  0x  =  -18 x - 20  =  0x  =  20

Because it is a positive integer, we should not take x = -18.  We take positive value for x.

If x  =  20,

then

y  =  360/20

y  =  18

Therefore the required positive integers are 20 and 18.

Verification :

Two natural numbers differ by 2.

20 - 18  =  2

their product is 360

20(18)  =  360

Problem 2 :

There are three consecutive positive integers such that the sum of the square of first and the product of the other two is 154. Find the integers.

Solution :

Let x, (x + 1) and (x + 2) be the first three consecutive integers

The sum of the squares of first and the product of the other two is 154

x2 + (x + 1)(x + 2)  =  154

x2 + x2 + 2x + 1x + 2  =  154

2x2 + 3x + 2  =  154

2x2 + 3x + 2 - 154  =  0

2x2 + 3x - 152  =  0

(2x + 19)(x - 8)  =  0

 2x + 19  =  0x  =  -19/2 x - 8  =  0x  =  8

Since it is positive integer, the value of x be 8.

Therefore three consecutive integers are 8, 9 and 10.

Verification :

The sum of the square of first and the product of the other two is 154.

82 + (9)(10)  =  154

64 + 90  =  154

154  =  154

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