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The sum of squares of two positive numbers is 100. If one number exceeds the other by 2, find the numbers.
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Let the numbers be $x, x + 2$
$x^2 + (x + 2)^2 = 100$ (1½ Mark)
simplifying we get
$2x^2 + 4x - 96 = 0$ or $x^2 + 2x - 48 = 0$ (1 Mark)
which gives $(x + 8) (x - 6) = 0$ (1/2 Mark)
$x = 6, - 8$ (1/2 Mark)
As $x > 0$ thus, numbers are 6, 8 (1/2 Mark)
$x^2 + (x + 2)^2 = 100$ (1½ Mark)
simplifying we get
$2x^2 + 4x - 96 = 0$ or $x^2 + 2x - 48 = 0$ (1 Mark)
which gives $(x + 8) (x - 6) = 0$ (1/2 Mark)
$x = 6, - 8$ (1/2 Mark)
As $x > 0$ thus, numbers are 6, 8 (1/2 Mark)