three consecutive natural number are such that the square of the middle number exceeds the the difference if the square of the other two by60. Assume the middle number to be x and form a quadratic equation satisfying the above statement.Hence,find the three number
@sepeario

Question

three consecutive natural number are such that the square of the middle number exceeds the the difference if the square of the other two by60. Assume the middle number to be x and form a quadratic equation satisfying the above statement.Hence,find the three number
@sepeario

sepeario · Answer

Say the three consecutive numbers are x-1, x and x+1. Hence the equation becomes $$x^2=[(x+1)^2-(x-1)^2]+60$$

sepeario · Answer

are you able to expand the equation above?

aaronandyson · Answer

No

sepeario · Answer

ok i think you should learn how to expand these equations in your free time but for now the equation is $$x^2=(x^2+2x+1)-(x^2-2x+1)+60$$ now by cancelling out what do we get?

sepeario · Answer

simplify the expression.

aaronandyson · Answer

I thought that only
But unable to move any further :(

aaronandyson · Answer

Is it like
x^2 - 4x - 60 = 0?
x = 10 & 6?

sepeario · Answer

Very good!

sepeario · Answer

Do be aware that it is 10 and -6, which you can see from the final equation.

aaronandyson · Answer