Question

we have 3 consecutive integers. the sum of their squares is 50. find the integers

Answer 1

Heya Friend !!

Let the three integers be x , x+1 , x+2

Given => x² + (x + 1)² + (x + 2)² = 50

=> x² + x² + 1 + 2x + x² + 4 + 4x = 50

=> 3x² + 6x + 5 = 50

=> 3x² + 6x = 45

=> 3x² + 6x - 45 = 0

=> 3x² -9x + 15x - 45 = 0

=> 3x(x - 3) + 15(x - 3) = 0

=> (x - 3)(3x + 15) = 0

=> x - 3 = 0 or 3x + 15 = 0

=> x = 3 or 3x = -15

=> x = 3 or x = -15/3

Neglecting the value of x as - 15/3 since it can't be negative .
=> x=3
=> x+1 = 3+1 = 4
=> x+2 = 3+2 =5
Hence , the numbers are 3 , 4 and 5 .

Verification :-
(3)^2 + (4)^2 +(5)^2 = 50
=> 9 + 16 + 25 =50
=> 50 = 50

Hence , LHS = RHS .

hope this helps :)

thank you :)

Answer 2

let 3 consecutive integers are x,(x+1),(x+2)
acc to question x²+(x+1)²+(x+2)2=50
x²+(x²+1²+2(x*1))+ x²+4+4x=50 (apply (a+b)²
x²+x²+x²+2x+4x+1+4=50
3x²+6x+5-50 =0(quadratic equation)
3x²+6x-45
3(x²+2x-15)=0
x²+2x-15=0
x²+(5x-3x)-15=0. (split middle term)
x(x+5)-3(x+5)
either x-3=0 or x+5= 0
x=3, x=-5
take positive value so,
x=3
so integers are 3,4,5