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

Given:

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

 To do:

We have to find the integers.

Solution:

Let the three consecutive integers be $x$, $x+1$ and $x+2$.

This implies,

The square of the number $x$ is $x^2$.

According to the question,

$x^2+(x+1)(x+2)=46$

$x^2+x^2+x+2x+2=46$

$2x^2+3x+2-46=0$

$2x^2+3x-44=0$

Solving for $x$ by factorization method, we get,

$2x^2+11x-8x-44=0$

$x(2x+11)-4(2x+11)=0$

$(2x+11)(x-4)=0$

$2x+11=0$ or $x-4=0$

$2x=-11$ or $x=4$

$x=\frac{-11}{2}$ or $x=4$

$\frac{-11}{2}$ is not an integer. Therefore, the required value of $x$ is $4$.

The required numbers are $4$, $5$ and $6$. 

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Updated on: 10-Oct-2022

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