Answers

The Brainliest Answer!
2016-01-02T16:06:22+05:30
Since the middle number of the three consecutive numbers is x, the other two numbers are x -1 and x +1.
According to the given condition, we have
x² = [(x +1)² -(x -1)²] +60
x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60
x² -4x -60 = 0 => (x -10)(x +6) = 0
x = 10 or -6
Since x is a natural number, we get x = 10
Hence the three numbers are 9, 10, 11
4 5 4
2016-01-02T17:41:11+05:30
Question: Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60.

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ANSWER:

There are three consecutive integers (same as, natural 
numbers).

let the smallest of the numbers be 
.

Values of 
 :

1st no. is x.
2nd no. is x+1.
3rd no. is x+1+1  =x+2 .

Condition: The square of the 2nd no. (middle number) is 60 more than the difference of the squares of other two numbers.

Solution:

(x+1)^{2}=[(x+2)^{2}-x^{2}]+60

=>    x^{2}+2x+1= (x^{2}+4x+4- x^{2})+60

=>    x^{2}+2x+1= x^{2}+4x+4- x^{2}+60

=>    x^{2}-x^{2}+x^{2}+2x-4x=4+60-1   

=>    (x^{2}-x^{2}+x^{2})+(2x-4x)=4+60-1  

=>   x^{2}-2x=64-1

=>    x^{2}-2x=63

=>   x= \sqrt{63}+2x

=>   x-2x= \sqrt{63}

=>   -x= \sqrt{63}

=>   x= \sqrt{-63}


Since,  \sqrt{-63} is an irrational number, so we will not further simplify it.


By putting the values of ,

1st no. is x= \sqrt{-63} .
2nd no. is x+1= \sqrt{-63}+1 .
3rd no. is x+2= \sqrt{-63}+2 .

ANSWER: The values of the three consecutive numbers:

1st no. is \sqrt{63} .
2nd no. is \sqrt{63}+1 .
3rd no. is \sqrt{63}+2 .

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