16
Three consecutive integers are such that sum of the square of second and product of other two is $161$. Find the three integers.
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Let the three numbers be $x, x+1$ and $x+2$ ($\frac{1}{2}$)
$(x + 1)^2 + x(x + 2) = 161$
$\Rightarrow x^2+2x-80 = 0$ (1)
$\Rightarrow (x+10)(x-8)=0$
$\therefore x= 8$ or $-10$ (1)
So, the numbers are $8, 9, 10$ or $-10, -9, -8$ ($\frac{1}{2}$)
$(x + 1)^2 + x(x + 2) = 161$
$\Rightarrow x^2+2x-80 = 0$ (1)
$\Rightarrow (x+10)(x-8)=0$
$\therefore x= 8$ or $-10$ (1)
So, the numbers are $8, 9, 10$ or $-10, -9, -8$ ($\frac{1}{2}$)