quadratic equation solution3

Quadratic Equation Solution3

In this page quadratic equation solution3 we are going to see solution of the word problems of the topic quadratic equation.

Question 5

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

Solution:

Let "x" and "y" are two natural numbers

it differs by 2

So, x - y = 2 ----- (1)

Their product is 360

So, x y = 360

y = 360/x ----- (2)

Now we are going to apply the value of y in the first equation.

x - (360/ x) = 2

(x² - 360)/x = 2

x² - 360 = 2 x

x² - 2 x - 360 = 0

x² - 20 x + 18 x - 360 = 0

x (x - 20) + 18 (x - 20) = 0

(x + 18) (x - 20) = 0

x + 18 = 0 x - 20 = 0

x = -18 x = 20

Since it is a positive integer we should not take x = -18. So let us take x = 20. Now we have to apply this value of x in the second equation in-order to get the value of y.

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

Question 6

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) are the first three consecutive integers

Here the sum of the square of first and the product of the other two is 154

x² + (x + 1) (x + 2) = 154

x² + x² + 2 x + 1 x + 2 = 154

2 x² + 3 x + 2 = 154

2 x² + 3 x + 2 - 154 = 0

2 x² + 3 x - 152 = 0

2 x² - 16 x + 19 x - 152 = 0

2 x (x - 8) + 19 (x - 8) = 0

(2 x + 19) (x - 8) = 0

2 x + 19 = 0 x - 8 = 0

2 x = -19 x = 8

x = -19/2

Since those are positive integer we should not take x = -19/2. So let us take the value 8 for x.

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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SBI!