Question

The sum of squares of three consecutive numbers is 77 find the numbers​

Answer 1

Answer:

4,5,6 are the three consecutive numbers. 16, 25 and 36 are their squares.

Step-by-step explanation:

Let the three consecutive numbers be x, (x+1), (x+2)

Now, the squares of these three numbers are  $x^{2} ,(x+1)^{2} ,(x+2)^{2}$

Sum = 77

∴by the problem ,

$x^{2} +(x+1)^{2}+(x+2)^{2} = 77\\x^{2} +(x^{2} +2x+1)+(x^{2}+4x +4) = 77\\x^{2} +x^{2} +2x+1+x^{2} +4x +4 = 77\\3x^{2} +6x+5 = 77\\3x^{2} +6x = 77-5\\3x^{2} + 6x = 72\\3x^{2} +6x-72= 0$

{Taking 3 common }

$x^{2} +2x- 24 = 0$

{By factorization}

$x^{2} +2x- 24 =0\\ x^{2} +6x-4x-24=0\\x(x+6)-4(x+6)=0\\(x+6)(x-4)=0$

Therfore,

$x= -6,4$

X can't be negetive

∴ $x =4\\x+1=5\\x+2=6$

The squares of the three consecutive numbers are 16, 25, 36

The three consecutive numbers whose sum is 77 are 4, 5, 6