Question

three consecutive positive integer are such that the sum of square of the first and product of other two is 199 find the integer

Answer 1

According to given condition,

( x + 1) {}^{2} + x( x + 2) = 199

x^2+2x+1+x^2+2x=199)[/tex [tex]2 x^2+4 x+1=199

2x^2+4x-198=0

{2x^2+22x-18x-198=0}

{2x(x+11)-18(x+11)=0}

(2x-18) (x+11)=0

{2x-18=0 \: or \: x+11=0}

x=18/2=9 \: or \: x=-11\\

( \: x = 9 )\: or \: (x = - 11)

According to the question, we want positive integer, so we cannot accept ( x = -11 )

So, x = 9

x = 9

x + 1 = 9+1 = 10

x + 2 = 9 + 2 = 11

Answer 2

Here is your answer..
Let the three consecutive integers be x , x+1 ,x +2

According to given condition,
$( x + 1) {}^{2} + x( x + 2) = 199$

$x^2+2x+1+x^2+2x=199)[/tex [tex]2 x^2+4 x+1=199$

$2x^2+4x-198=0$

${2x^2+22x-18x-198=0}$

${2x(x+11)-18(x+11)=0}$

$(2x-18) (x+11)=0$

${2x-18=0 \: or \: x+11=0}$

$x=18/2=9 \: or \: x=-11$
$( \: x = 9 )\: or \: (x = - 11)$
According to the question, we want positive integer, so we cannot accept ( x = -11 )
So, x = 9
x = 9
x + 1 = 9+1 = 10
x + 2 = 9 + 2 = 11
So the three consecutive integers are 9 ,10 ,11
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THIS IS THE ANSWER.
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