Question 859547


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^2 = (x+2)+10}}}
You have a quadratic equation which you need to simplify and put in standard form.
{{{x^2 = x + 12}}}
{{{ x^2 - x - 12 = 0 }}}


Now you solve for x - break the middle term into two parts so that you can factor the left side.
{{{x^2 -4x + 3x - 12 = 0}}}
{{{x*(x - 4) + 3*(x - 4) = 0}}}
{{{(x + 3)*(x - 4) = 0}}}


Two options:
{{{x + 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?
{{{4^2 - 6 = 16 - 6 = 10}}}
All good :)