Question

the sum of the squares of two consecative natural number is 313 find the numbers

Answer 1

Answer:

let the numbers are x and x+1

so, x^2+(x+1)^2=313

=> x^2+x^2+2x+1=313

=> 2x^2+2x+1-313=0

=> 2x^2+2x-312=0

=> x^2+x--156=0

=> x^2+13x-12x-156=0

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

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

x= -13 , x = 12

so the numbers are 12 and 13

Answer 2

To find : two consecutive numbers

Given : sum of the squares of two consecutive natural number is $313$ .

Solution :  

• As per given data we know that sum of the squares of two consecutive natural number is $313$ .
• We know that two consecutive numbers differ by $1$ .
• Therefore , let $x$ and $x + 1$ be two consecutive numbers .
• We represent given condition as , $x^{2} + (x+1)^{2} = 313$
• Applying identity $(a + b )^{2} = a^{2} + 2ab + b^{2}$ to the term $(x + 1)^{2}$ .

• $x^{2} + (x+1)^{2} = 313$  

      $x^{2} + x^{2} +2x + 1 =313 \\2x^{2} + 2x - 312 =0 \\2x^{2} -24x + 26x -312 =0 \\2x (x -12 ) +26 (x -12 ) =0 \\(x -12 )(x + 26 )\\x = 12 \ or \ x = -26$

• As we know that natural numbers are the all positive integers from $1$ to infinity .
• Hence we take value of $x$ as $12$ .
• Now , consecutive numbers are :

       $x = 12\\x + 1 =12 + 1 = 13$

• Verifying with given condition :

         $12^{2} + 13^{2} = 144 + 169 = 313$

        Hence , verified.

Hence ,  two consecutive numbers are $12$ and $13$.