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
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
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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