SOLUTION: find two consectutive postitive integers if the sum of their squares is 61 2n=61

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Question 221401: find two consectutive postitive integers if the sum of their squares is 61

2n=61
Found 2 solutions by drj, MathTherapy:
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find two consecutive positive integers if the sum of their squares is 61.

Step 1. Let n be a positive integer.

Step 2. Let n+1 be the next positive and consecutive number.

Step 3. Let n%5E2%2B%28n%2B1%29%5E2=61 since the sum of their squares is 61.

Step 4. Solving yields the following steps

n%5E2%2Bn%5E2%2B2n%2B1=61

2n%5E2%2B2n%2B1=61

Subtract 61 from both sides of the equation to get a quadratic equation.

2n%5E2%2B2n%2B1-61=61-61

2n%5E2%2B2n-60

Step 5. To solve, use the quadratic formula given as

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=2, b=2 and c=-60

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B2x%2B-60+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282%29%5E2-4%2A2%2A-60=484.

Discriminant d=484 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+484+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+484+%29%29%2F2%5C2+=+5
x%5B2%5D+=+%28-%282%29-sqrt%28+484+%29%29%2F2%5C2+=+-6

Quadratic expression 2x%5E2%2B2x%2B-60 can be factored:
2x%5E2%2B2x%2B-60+=+2%28x-5%29%2A%28x--6%29
Again, the answer is: 5, -6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B2%2Ax%2B-60+%29



Select the positive solution n=5 then n+1=6 and 5%5E2%2B6%5E2=61...a true statement.

Step 6. ANSWER: The consecutive positive integers are 5 and 6

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J






Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!
find two consectutive postitive integers if the sum of their squares is 61

Let the 1st integer be F

Then the 2nd integer is: F + 1

Since the sum of their squares is 61, then we'll have: F%5E2+%2B+%28F+%2B+1%29%5E2+=+61

F%5E2+%2B+F%5E2+%2B+2F+%2B+1+=+60

2F%5E2+%2B+2F+-+60+=+0

2%28F%5E2+%2B+F+-+30%29+=+2%280%29 ------ Factoring out the GCF, 2

F%5E2+%2B+F+-+30+=+0

(F + 6)(F - 5) = 0

(F + 6) = 0, or (F - 5) = 0

Since we're looking for positive integers, we ignore (F + 6) = 0, as this will give us a negative value (- 6) for F

Therefore, F, or the 1st of the 2 integers = 5

This means that the 2 positive integers are: highlight_green%285%29 and highlight_green%286%29