Question

The sum of the squares of two consecutive natural numbers is 313. find the numbers

Answer 1

let the numbers be x,x+1
so, x2+(x+1)^2=313
x2+x2+2x+1=313
2x^2+2x=312
(/2) x2+x=156
by quadratic formula,
x= -b+- √b2-4ac/2
= -1 +- √1*1 -4*1*-156 /2
= -1+- √1+624/2
= -1+- √625/2
=-1+-25/2
=-1+25/2, -1-25/2
=24/2 ,-26/2
= 12, -13
given nos are natural . so x= 12 and x+1=13

Answer 2

Answer:

Given,

sum of squares of two consecutive numbers = 313

to find : the two numbers

Explanation:

let the two numbers be x, x+1

the sum of their squares is 313

${x}^{2} + {(x + 1)}^{2} = 313 \\ {x}^{2} + {x}^{2} + 2x + 1 = 313 \\ 2 {x}^{2} + 2x = 312$

$2( {x}^{2} + x) = 312 \\ {x}^{2} + x = 156$

${x}^{2} + x - 156 = 0$

now we can solve this quadratic equation by factorisation,

${x}^{2} + 13x - 12x - 156 = 0 \\ x(x + 13) - 12(x + 13) = 0 \\ (x - 12)(x + 13) = 0$

x= +12 and -13

the numbers are natural so we have to ignore the negative numbers

x= 12, x+1= 13