The sum of the squares of two consecutive real numbers is 61. Find the numbers.

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We are given that the sum of the squares of two consecutive numbers is equal to 61. Let the numbers be x and x+1.

So x^2 + (x+1)^2 = 61

=> x^2 + x^2 + 1 + 2x = 61

=> 2x^2 + 2x - 60 = 0

=> x^2 + x - 30 =0

=> x^2 + 6x - 5x - 30 =0

=> x(x+6) -5(x+6) =0

=>(x-5)*(x+6) =0

We can have x = 5 and x = -6

Therefore the numbers can be 5 and 6 or -6 and -5.

The required result is (5, 6)  and (-6,-5).

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Let the first number be x

Then the next number will be x + 1.

Given that the sum of the squares  is 61

Then :

x^2 + ( x+1)^2 = 61

==> x^2 + x^2 + 2x + 1 = 61

We will combine like terms:

==> 2x^2 + 2x + 1 - 61 = 0

==> 2x^2 + 2x - 60 = 0

\We will divide by 2:

==> x^2 + x - 30 = 0

Now we will factor:

==> ( x - 6)( x+ 5)

==> x 1= 6  ==>  the second number = 5

==> x2= -5 ==> the second number = -6

Then the numbers are:

( -5, -6 )  OR  ( 5, 6)

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