76
Three consecutive positive integers are such that sum of square of the first and the product of the other two is $67$, find the integers.
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Let the consecutive integers be $x, (x + 1)$ and $(x + 2)$ (1 Mark)
$x^2 + (x + 1)(x + 2) = 67$ (1 Mark)
$\Rightarrow 2x^2 + 3x - 65 = 0$ (1 Mark)
$\Rightarrow (x - 5)(2x + 13) = 0$ (1 Mark)
$\Rightarrow x = 5$ or $x = -\frac{13}{2}$ (1 Mark)
neglecting $x = -\frac{13}{2}$, as $x$ is a positive integer. ($\frac{1}{2}$ Mark)
So, $x = 5$
Thus, integers are $5, 6$ and $7$ ($\frac{1}{2}$ Mark)
$x^2 + (x + 1)(x + 2) = 67$ (1 Mark)
$\Rightarrow 2x^2 + 3x - 65 = 0$ (1 Mark)
$\Rightarrow (x - 5)(2x + 13) = 0$ (1 Mark)
$\Rightarrow x = 5$ or $x = -\frac{13}{2}$ (1 Mark)
neglecting $x = -\frac{13}{2}$, as $x$ is a positive integer. ($\frac{1}{2}$ Mark)
So, $x = 5$
Thus, integers are $5, 6$ and $7$ ($\frac{1}{2}$ Mark)