SOLUTION: find three consecutive integers such that the sum of the squares of the second and the third exceeds the square of the first by 21.

Question 162007: find three consecutive integers such that the sum of the squares of the second and the third exceeds the square of the first by 21.
Found 4 solutions by SAT Math Tutor, Earlsdon, eperette, josmiceli:
Answer by SAT Math Tutor(36)   (Show Source): You can put this solution on YOUR website!
Ok, the first thing to do is set up your variables. We will choose 3 consecutive integers using just 1 variable n:
n
n + 1
n + 2
Then, set up the equation:

And solve:

Then, use the quadratic equation:

This gives you an answer of n = 2, -8
So there are 2 solutions:
2, 3, 4
AND
-8, -7, -6
Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
Let the three consecutive integers be: x, (x+1), and (x+2)
"...the sum of the squares of the second and third exceed the square of the first by 21." Simplify this and solve for x.
Combine like-terms.
Subtract from both sides.
Subtract 21 from both sides.
Solve this quadratic equation by factoring.
Apply the zero product rule.
or so...
or
There are two answers to this problem:
x = 2
(x+1) = 3
(x+2) = 4 and...
x = -8
x+1 = -7
x+2 = -6
Answer by eperette(173)   (Show Source): You can put this solution on YOUR website!

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Call the cosecutive integers ,,
Given:

Either

or

Use the positive result

The numbers are 2,3,and 4
check answer:

OK
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