Question

13. Three consecutive natural numbers are such
that the square of the middle number exceeds
the difference of the squares of the other two
by 60
sume the middle number to be and form​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let the three consecutive natural numbers be x,x+1, x+2.

Given that Square of the middle number exceeds the difference of the squares of the other two by 60.

(x + 1)^2 = (x + 2)^2 - (x)^2 + 60

x^2 + 1 + 2x = x^2 + 4 + 4x - x^2 + 60

x^2 + 1 + 2x = 4x+ 64

x^2 = 4x + 64 - 2x - 1

x^2 = 2x + 63

x^2 - 2x - 63 = 0

x^2 - 9x + 7x - 63 = 0

x(x - 9) + 7(x - 9) = 0

(x - 9)(x + 7) = 0

x = 9,-7.

x value cannot be -ve, so, x = 9.

Then,

x + 1 = 10

x + 2 = 11.

Therefore the three natural numbers are 9,10,11.

Verification:

(x + 1)^2 - (x + 2)^2 + x^2 = 60

10^2 - 11^2 + 9^2 = 60

100 - 121 + 81 = 60

-21 + 81 = 60

60 = 60.

Hope this helps!