# 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?

Relevance
• 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

• Let x-1, x be the consecutive integers, then

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

=>

x=?

You should find it out by yourself.

• 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

• 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...

• 2x^2 + 2x = 1512

x(x + 1) = 756 = 27*28

• 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.

• 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

• 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