Find 2 consecutive integers such that the sum of their squares is 113.?

Question

Yes, I know the answer is 7 and 8, or -7 and -8, but I don't know how to do the work. This is where I get stuck...

x^2 + (x+1)^2 = 113

x^2 + x^2 +1 = 113

This SHOULD involve polynomial equations....

Answer 1

You screwed up in the expansion of (x + 1)^2. Let's FOIL to see what it actually should be:

(x + 1)^2
==> rewrite as two binomials
(x + 1)(x + 1)
==> FOIL: first, outside, inside, last
x^2 + x + x + 1
==> simplify
x^2 + 2x + 1

So, we'll plug this in:

x^2 + (x^2 + 2x + 1) = 113
==> combine like terms
2x^2 + 2x + 1 = 113
==> subtract 113 from both sides
2x^2 + 2x - 112 = 0
==> divide everything by 2
x^2 + x - 56 = 0
==> factor
(x + 8)(x - 7) = 0
==> two answers
x = -8 or x = 7
==> these are the first numbers of your two possible sets
ANSWER: -8 and -7 OR 7 and 8

Answer 2

You forgot some 2's. Your second equation should read
2x²+2x+1 = 113.
So
2x²+2x-112 = 0
x²+x-56 = 0.
(x-7)(x+8) = 0
So either x = 7, x+1 = 8
or
x = -8, x+1 = -7.
In either case, the sum of the squares is 49+64 = 113.

Answer 3

Call them X and Y.

They're consecutive, so Y = X + 1
Their sum is 113, so X^2 + Y^2 = 113

Plug the first equation into the second.

X^2 + (X+1)^2 = 113
X^2 + (X^2 + 2X + 1) = 113
2*(X^2) + 2X - 112 = 0

This is of the form A*(X^2) + B*X + C = 0
A = 2 ; B = 2 ; C = -112

Use the quadratic formula.

X = (-B +/- sqrt(B^2 - 4*A*C))/2A

X = (-2 +/- sqrt(4 - 4(2)(-112)))/4
X = (-2 +/- sqrt(4 + 896)/4
X = (-2 +/- sqrt(900))/4
X = (-2 +/- 30)/4

(-2 + 30)/4 = 28/4 or 7 for X
(-2 - 30)/4 = -32/4 or -8 for X

X = 7 means Y = 8
X = -8 means Y = -7

Whew!

Answer 4

7^2=49+(7+1)^2=113, because 8^2=64

49+64=113