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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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