Respuesta :
Answer:
The three consecutive natural numbers are 9,10,11
Step-by-step explanation:
Step 1 : -
Let x be the number
Given three consecutive natural numbers are x, x+1,x+2
Given data are 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
that is [tex](x+1)^2 = (x+2)^2 - x^2 +60{\tex]
step 2:-
on simplification on both sides are , we get
by using [tex](a+b)^2 = a^2+2 a b+b^2[\tex]
[tex]x^2+2 x+1 = x^2+4 x+4-x^2+60[\tex]
cancelling x^2 terms and simplify
[tex] x^2 -2 x-63=0[\tex]
now finding factors of [tex] 63 = 9 X 7[\tex]
[tex] x^2 - 9 x+7 x -63=0 [\tex]
[tex] x(x-9)+7(x-90 =0 [\tex]
[tex]( x+7)(x-9) =0 [\tex]
here x= -7 is not an natural number
so we have to take x=9
therefore the three consecutive natural numbers are
x,x+1,x+2
The three consecutive natural numbers are 9 , 10, 11
Answer:
three consecutive natural numbers are
5,6,7
Step-by-step explanation:
Let x, x+1 and x+2 are the three consecutive natural numbers
middle number is x+1
square of middle number is [tex](x+1)^2=x^2+2x+1[/tex]
difference of square of other two numbers is
[tex]x^2-(x+2)^2= x^2-x^2-4x-4=-4x-4[/tex]
the square of the middle number exceeds the difference of the squares of the other two numbers by 60
So [tex]x^2+2x+1=-4x-4+60[/tex]
[tex]x^2+2x+1+4x+4-60=0[/tex]
[tex]x^2+6x-55=0[/tex]
[tex](x+11)(x-5)=0[/tex]
[tex]x+11=0, x=-11[/tex]
[tex]x-5=0, x=5[/tex]
we take only natural number so x=5
three consecutive natural numbers are
5,6,7
