Question

the sum of the squares of two consecutive natural numbers in 313.Find the numbers

Answer 1

Answer:

let one no. be x and the other no. be x+1.

square both sides and add together to equal 313.

x2+ (x+1)2=313

x2+x2+2x+1=313

2x2+2x-312=0

x2+x-156=0

(x-12)(x+13)=0

x=12and x+1=-13

neglecting (-) sign

so one no. is 12 and the other no. is 13

Step-by-step explanation:

Answer 2

Answer:

12, 13

Step-by-step explanation:

$\text{Let the 2 consecutive numbers be x and (x+1)}\\\\\text{Then, according to the question,}\\\\x^2 + (x+1)^2 = 313\\\\\implies x^2 + (x^2 + 1^2 + 2(x)(1)) = 313\\\\\implies x^2 + x^2 + 1 + 2x = 313\\\\\implies 2x^2 + 2x + 1 = 313\\\\\implies 2x^2 + 2x + 1 - 313 = 0\\\\\implies 2x^2 + 2x + 1 -312 = 0\\\\\text{Dividing both sides by 2}\\\\\dfrac{1}{2} \times (2x^2+2x-312) = \dfrac{1}{2} \times 0\\\\\\\implies x^2 + x - 156 = 0$

$(x-12)(x+13) = 0\\\\\implies x = 12 \: and \: x + 1 = - 13\\\\\text{Neglecting (-) sign,}\\\\\text{One number is x = 12 and the other number is x+1 = 12+1 = 13}$

Hope it helps!