Question

The sum of the squares of two consecutive integers is 145. Find the integers.

Answer 1

Let us call the numbers:n

and n+1

we get (from our condition) that:

(n)2+(n+1)2=145

rearrange and solve for

n :n2+n2+2n+1−145=0

2n2+2n−144=0

use the Quadratic Formula:

n1 , 2= −2 ± √4+1152 / 4 = −2±34/4

so ve get two values:

n1=−9

n2=8

we chose the positive one so that our numbers will be:

n=8

and

n+1=9

Hope this would help......

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Answer 2

Given:

Sum of the squares of two consecutive integers = 145

To find:

The integers

Solution:

Let the first integer be = n

Let the second integer be = n + 1

Therefore,

n² + (n +1)²= 145

Solving -

n² + n² + 2n + 1 = 145

2n²+ 2n - 144 = 0

n² + n - 72 = 0

n² + 9n - 8n - 72 = 0

n(n + 9) - 8(n + 9) = 0

(n - 8)(n + 9) = 0

Thus, n can be 8 or -9

Answer : The two numbers are (8 , 9) or ( -8 , -9).

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