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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