Question
Find two consecutive odd positive integers, sum of whose squares is $970$.
✓
Answer
Let two consecutive positive integers be x and $x + 2$
$A.T.Q., (x)^2 + (x + 2)^2 = 970$
$\Rightarrow x^2 + x^2 + 4x + 4 - 970 = 0$
$\Rightarrow 2x^2 + 4x - 966 = 0$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x - 21)(x + 23) = 0$
Either $x - 21 = 0 or x + 23 = 0$
$x = 21 or x = -23$ (rejected being -ve)
As integers should be +ve
$x = 21$ and $x + 2 = 21 + 2 = 23$
Hence integers are $21, 23$
$A.T.Q., (x)^2 + (x + 2)^2 = 970$
$\Rightarrow x^2 + x^2 + 4x + 4 - 970 = 0$
$\Rightarrow 2x^2 + 4x - 966 = 0$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x - 21)(x + 23) = 0$
Either $x - 21 = 0 or x + 23 = 0$
$x = 21 or x = -23$ (rejected being -ve)
As integers should be +ve
$x = 21$ and $x + 2 = 21 + 2 = 23$
Hence integers are $21, 23$
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