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

aaronandyson · Answer

@butterflydreamer

aaronandyson · Answer

Is it 9,10,11

aaronandyson · Answer

@mathslover 
app hamri help kar skte hai? :)

mathslover · Answer

Yup, it's 9 , 10, 11

anonymous · Answer

Yup

aaronandyson · Answer

Although I got the answer let's calculate it again ?please?

mathslover · Answer

Let the consecutive numbers be : $x-1 , x , x + 1 $ 

Given that : $x^2$ exceeds the difference of the other two numbers by 60.

Means : $x^2 - \left( (x+1)^2 - (x-1)^2 \right) = 60$ 

Right?

aaronandyson · Answer

Yes.

mathslover · Answer

$x^2 - \left( (x+1 + x-1) (x+1 -(x-1))\right) = 60 \
x^2 - \left((2x)(2) \right) = 60$ 

Right?

aaronandyson · Answer

Yes.

mathslover · Answer

I just used $a^2 - b^2 = (a+b)(a-b)$ identity there.

Okay, you've : $x^2 - 4x = 60$ or $x^2 - 4x - 60 = 0 $ 

I can write it as : $(x - 10)(x + 6) = 0 $ 

So, roots are : x = 10 or x = - 6

Since negative numbers can not be natural number or vice versa.

Therefore, x = 10 

And x - 1 = 10 - 1 = 9 and x + 1 = 11

So, numbers are : 9,10,11

aaronandyson · Answer

Okay,I need a little with like more 5 problem can you stick around for awhile?
And thanks:)

mathslover · Answer

You're welcome Aaron! I may have to go right now. Hopefully any other will help you. Try to stick with basics and you'll get the instinct required to do questions very soon. Take Care.