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Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.
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Let the three consecutive positive integers be $x, x + 1, x + 2$ [$1$ mark]
A.T.Q. $x^2 + (x + 1) (x + 2) = 67$ [$1$ mark]
$\Rightarrow 2x^2 + 3x - 65 = 0$ [$1$ mark]
$\Rightarrow (2x + 13) (x - 5) = 0$ [$1$ mark]
$\Rightarrow x = 5, x = -\frac{13}{2}$ (rejected)
So the three consecutive positive integers are 5, 6 and 7 [$1$ mark]
A.T.Q. $x^2 + (x + 1) (x + 2) = 67$ [$1$ mark]
$\Rightarrow 2x^2 + 3x - 65 = 0$ [$1$ mark]
$\Rightarrow (2x + 13) (x - 5) = 0$ [$1$ mark]
$\Rightarrow x = 5, x = -\frac{13}{2}$ (rejected)
So the three consecutive positive integers are 5, 6 and 7 [$1$ mark]