# Need help with a question on Quadratic equations in math?

Two consecutive integers are squared. The sum of these squares is 1513.

What are the integers?

### 8 Answers

- lenpol7Lv 72 months ago
let the integers by 'n' & 'n+1'

Squaring

n^2 + ( n + 1)^2 =

n^2 + n^2 + 2n + 1 and it equals = 1513

2n^2 + 2n + 1 = 1513

2n^2 + 2n = 1512

Factor out '2'

n^2 + n -756 = 0

This factors to

( n + 28)(n - 27) = 0

Hence n= 27

& n + 1 = 28

- ?Lv 72 months ago
Let x-1, x be the consecutive integers, then

(x-1)^2+x^2=1513

=>

x=?

You should find it out by yourself.

- RRLv 62 months ago
n² + (n + 1)² = 1513

n² + n² +2n + 1 = 1513

2n² + 2n -1512 = 0

n² + n - 756 = 0

(n + 28)(n - 27) = 0

n = 27 or n = - 28

ANS 27 and 28 or -27 and -28

- Engr. RonaldLv 72 months ago
n^2 + (n + 1)^2 = 1513

n^2 + n^2 + 2n + 1 = 1513

2n^2 + 2n + 1 = 1513

2n^2 + 2n + 1 - 1513 = 0

2n^2 + 2n - 1512 = 0

n^2 + n - 756 = 0

(n + 28)(n - 27) = 0

n =- 28, n = 27

the integers are- 28 and 27...

- What do you think of the answers? You can sign in to give your opinion on the answer.
- ?Lv 72 months ago
Two consecutive integers are squared.

The sum of these squares is 1513.

x^2 + (x + 1)^2 = 1513

2x^2 + 2x - 1512 = 0

x^2 + x - 756 = 0

(x - 27) (x + 28) = 0The integers are 27 and 28.

- billrussell42Lv 72 months ago
call them x and x+1

sum of squares is

x² + (x+1)² = 1513

x² + x² + 2x + 1 – 1513 = 0

2x² + 2x – 1512 = 0

x = 27, – 28 (using a quadratic solver)

so the integers are 27, 28

or –27, –28

- ?Lv 72 months ago
Consecutive integers differ by 1. If you call the smaller number x, then the other is x+1.

We can thus translate the statement into the equation

x^2 + (x+1)^2 = 1513

Expand the (x+1)^2 term, then collect like terms to make an equation of the form Ax^2 + Bx + C = 0.

Then either factor or use the quadratic equation to solve for x.

You will get two solutions, one positive and one negative, whose absolute values differ by 1. This is because positive and negative numbers have the same square. For example, 2^2 + 3^2 = (-2)^2 + (-3)^2