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
since the sum of their squares is 61.
Step 4. Solving yields the following steps
Subtract 61 from both sides of the equation to get a quadratic equation.
Step 5. To solve, use the quadratic formula given as
where a=2, b=2 and c=-60
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=484 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 5, -6.
Here's your graph:
 |
Select the positive solution n=5 then n+1=6 and
...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
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:
------ Factoring out the GCF, 2
(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 =
This means that the 2 positive integers are:
and
