Question

There are three consecutive positive integers wherein the sum of the square of the first integer and the product of the other twobis 29,find the integers

Answer 1

Given : There are three consecutive positive integers where in the sum of the square of the first integer and the product of the other two is 29

To find:   the integers

Solution:

Let say three consecutive positive integers  are n , n +1 , n + 2

sum of the square of the first integer and the product of the other two is 29

=> n² + (n +1 )(n + 2)  = 29

=> n² + n² + 3n + 2 = 29

=> 2n² + 3n - 27 = 0

=> 2n² - 6n +9n - 27 = 0

=> 2n(n  - 3) + 9(n - 3) = 0

=> (n - 3)(2n + 9) = 0

=> n = 3  , n = - 9/2

n is a positive integer hence -9/2 is not possible

so n = 3

Hence integers are  3 , 4 , 5

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Answer 2

Given,

Sum of the square of the first integer and the product of the other two is 29.

Solution,

Consider the integers are $\[x - 2,x - 1,x\]$.

According to the given condition,

$\[\begin{array}{l}{\left( {x - 2} \right)^2} + x\left( {x - 1} \right) = 29\\ \Rightarrow {x^2} - 4x + 4 + {x^2} - x = 29\\ \Rightarrow 2{x^2} - 5x - 25 = 0\\ \Rightarrow 2{x^2} - 10x + 5x - 25 = 0\\ \Rightarrow 2x\left( {x - 5} \right) + 5\left( {x - 5} \right) = 0\\ \Rightarrow \left( {x - 5} \right)\left( {2x + 5} \right) = 0\\ \Rightarrow x = 5,\frac{{ - 5}}{2}\end{array}\]$

Integers,

$\[\begin{array}{l} = x - 2,x - 1,x\\ = 5 - 2,5 - 1,5\\ = 3,4,5\end{array}\]$

Hence the integers are $\[3,4,5\]$