Let one of the numbers be x.

Then the next consecutive number is (x+1)

Given that the sum of the squares is 145.

==> x^2 + (x+1)^2 = 145

==> x^2 + x^2 + 2x + 1 = 145

==> 2x^2 + 2x - 144 = 0

Now let us divide by 2.

==> x^2 + x - 72 = 0

Now we will factor.

==> (x +9) ( x-8) = 0

==> x1= -9 ==> x1+1 = -8

==> x2= 8 ==> x2+1 = 9

**Then the numbers are (8 and 9) or ( -8 and -9)**

Posted on

The sum of the squares of two consecutive numbers is 145.

Let the numbers be N and N + 1

Now we have N^2 + (N +1)^2 = 145

=> N^2 + N^2 + 2N + 1 = 145

=> 2N^2 + 2N - 144 = 0

=> N^2 + N - 72 = 0

=> N^2 + 9N - 8N - 72 = 0

=> N(N + 9) - 8(N + 9) = 0

=> (N - 8)(N + 9) = 0

N can be 8 or -9

**So the two numbers are (8 , 9) or ( -8 , -9)**.

Posted on

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