There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?

Given:

There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. 

 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,

$x^2+(x+1)(x+2)=154$

$x^2+x^2+x+2x+2=154$

$2x^2+3x+2-154=0$

$2x^2+3x-152=0$

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

$2x^2+19x-16x-152=0$

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

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

$2x+19=0$ or $x-8=0$

$2x=-19$ or $x=8$

$x=\frac{-19}{2}$ or $x=8$

$\frac{-19}{2}$ is not an integer. Therefore, the required value of $x$ is $8$.

The required numbers are $8$, $9$ and $10$.

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Updated on: 10-Oct-2022

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