Question
Find two consecutive numbers whose squares have the sum 85.
✓
Answer
Let two consecutive numbers be X and (x + 1)
Then according to question
$x^2 + (x + 1)^2 = 85$
$x^2 + x^2 + 2x + 1 = 85$
$2x^2 + 2x - 85 + 1 = 0$
$2x^2 + 2x - 84 = 0$
$x^2 + x - 42 = 0$
$x^2 + 7x - 6x - 42 = 0$
$x(x + 7) - 6(x + 7) = 0$
$(x + 7)(x - 6) = 0$
$(x + 7) = 0$
$x = -7$
$or (x - 6) = 0$
$x = 6$
Since, x being a number,
Therefore,
When x = -7 then
x + 1 = -7 + 1
x + 1 = -6
And when x = 6 then
x + 1 = 6 + 1
x + 1 = 7
Thus, two consecutive number be either 6, 7 or -6, -7
Then according to question
$x^2 + (x + 1)^2 = 85$
$x^2 + x^2 + 2x + 1 = 85$
$2x^2 + 2x - 85 + 1 = 0$
$2x^2 + 2x - 84 = 0$
$x^2 + x - 42 = 0$
$x^2 + 7x - 6x - 42 = 0$
$x(x + 7) - 6(x + 7) = 0$
$(x + 7)(x - 6) = 0$
$(x + 7) = 0$
$x = -7$
$or (x - 6) = 0$
$x = 6$
Since, x being a number,
Therefore,
When x = -7 then
x + 1 = -7 + 1
x + 1 = -6
And when x = 6 then
x + 1 = 6 + 1
x + 1 = 7
Thus, two consecutive number be either 6, 7 or -6, -7
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