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

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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