SOLUTION: It would be greatly appreciated if you would be able to help me with: Find three consecutive, positive integers such that the square of the smallest exceeds the largest by 10.

Algebra ->  Human-and-algebraic-language -> SOLUTION: It would be greatly appreciated if you would be able to help me with: Find three consecutive, positive integers such that the square of the smallest exceeds the largest by 10.      Log On


   

Question 859547: It would be greatly appreciated if you would be able to help me with:
Find three consecutive, positive integers such that the square of the smallest exceeds the largest by 10.
I've attempted to solve this by doing:
(x+1)(x+2)(x+3)=10
As I'm not very familiar with verbal algebraic terms, I have trouble breaking down world problems.
Thanks for your help!
Answer by Awesom3guy(31) About Me  (Show Source):
You can put this solution on YOUR website!

You are looking for three consecutive numbers:
x, x+1, and x+2.
There's no need to make it x+1 on the first one.

The square of the smallest number exceeds the largest number by 10.
So the square of the smallest number is larger than the biggest number by 10.
Now you just need to put that in math language. It's REALLY not hard.
x%5E2+=+%28x%2B2%29%2B10
You have a quadratic equation which you need to simplify and put in standard form.
x%5E2+=+x+%2B+12
+x%5E2+-+x+-+12+=+0+

Now you solve for x - break the middle term into two parts so that you can factor the left side.
x%5E2+-4x+%2B+3x+-+12+=+0
x%2A%28x+-+4%29+%2B+3%2A%28x+-+4%29+=+0
%28x+%2B+3%29%2A%28x+-+4%29+=+0

Two options:
x+%2B+3+=+0, which gives x = -3. This obviously isn't the solution as we are looking for two consecutive POSITIVE integers.
x+-+4+=+0, which gives x = 4.

Now that you have the first number, you know that the other two are 5 and 6.
So the three integers are 4, 5 and 6.

Also check - does the square of 4 exceed 6 by 10?
4%5E2+-+6+=+16+-+6+=+10
All good :)