SOLUTION: find two consecutive positve intergers such that the sum of there squares is 41

Question 92505: find two consecutive positve intergers such that the sum of there squares is
41
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Translate the word problem:
"two consecutive positive integers such that the sum of their squares is 41"

Foil

Subtract 41 from both sides

Combine like terms

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=2, b=2, and c=-40

Square 2 to get 4

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our possible solutions are:
or

Since we only care about the positive numbers, our only solution is

So if x=4 then...

the 2nd number is 5

So the pair of numbers is 4 and 5

Check:

plug in the pair of numbers

Square each term

Combine like terms. Since the equation is true, our answer is verified.

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