Question

the sum of the squares of three consecutive natural number is 77. find the number

Answer 1

hey!
#ur Ans
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let the three consecutive number are :

X, (X+1) ,(X+2)

now sum of square

x^2 +(X+1)^2 +(X+2)^2 = 77

x^2 +x^2 +1 +2x +x^2 +4 +4x = 77

3x^2 +6x +5 = 77

3x^2 +6x = 77-5

3x^2 +6x -72=0

or

x^2 +2x -24 =0

nOw factorise

x^2 +6x -4x -24 =0

x(X+6) -4(X+6) =0

(x-4) (X+6) =0

X = 4 , -6

X can't be negative

so

three consecutive number are

X = 4
X+1 = 5
X+2 = 6

☺☺:-)✌

Answer 2

Heya!! Here is your answer friend ⤵⤵⤵
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⚫Let the three consecutive natural numbers be X , X+1 and X+2

Now According to the given question we have ⤵

➡
$x {?}^{2} + x {?}^{2} + 1 + 2x + x {?}^{2} + 4 + 4x = 77\\ 3x {?}^{2} + 6x + 5 - 77 = 0 \\ 3x {?}^{2} + 6x - 72 = 0 \\ taking \: three \: common \\ x { }^{2} + 2x - 24 \\ using \: middle \: term \: splitting \\ x {?}^{2} + 6x - 4x - 24 = 0 \\ x(x + 6) - 4(x + 6) \\ (x - 4)(x + 6) \\ x = 4 \\ x = - 6 \\ \\ therfore \: the \: numbers \: are \: \\ 4 \\ 5 \\ 6$
Hope it helps you ✌✌