Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.


Given: Three consecutive positive integers and Sum of the square of the first and product of the other two is 46. 
To do: To find out the integers.
Solution:
Let the three integers are $x-1,\ x,\ x+1$.
According to the given condition
$( x-1)^{2} +x( x+1) =46$
$\Rightarrow x^{2} -2x+1+x^{2} +x=46$
$\Rightarrow 2x^{2} -x+1=46$
$\Rightarrow 2x^{2} -x-45=0$
$\Rightarrow ( x-5) \ ( 2x+9) \ =0$
$\Rightarrow x=5\ or\ x=-\frac{9}{2}$
Here $x=5$ only because it is given that x is a positive integer.
$\therefore$ First positive integer$=x-1=5-1=4$
Second positive integer$=x=5$
And the third positive integer$=x+1=5+1=6$
Thus, the required integers are 4, 5 and 6.

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Updated on: 10-Oct-2022

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