Question

Question #852f1

Answer 1

`13` `&` `14`

Explanation:

Let the two consecutive positive integers be `x` `&` `x+1.`

`x^2+(x+1)^2=365`

We know that `(a+b)^2=a^2+2ab+b^2.`So the equation becomes

`=>x^2+(x^2+2*x*1+1^2)=365`

`=>x^2+x^2+2*x*1+1^2=365`

`=>x^2+x^2+2x+1=365`

`=>2x^2+2x+1=365`

`=>2x^2+2x+1color(red)(-365)=365color(red)(-365)`

`=>2x^2+2x-364=0`

Apply Quadratic Formula, that is `x=(-b+-sqrt(b^2-4ac))/(2a).` Here

`a=2, b=2, c=-364`

`=>x=(-2+-sqrt(2^2-(4*2*(-364))))/(2*2)`

`=>x=(-2+-sqrt(4-(8*(-364))))/(4)`

`=>x=(-2+-sqrt(4-(-2912)))/4`

`=>x=(-2+-sqrt(4+2912))/4`

`=>x=(-2+-sqrt(2916))/4`

`=>x=(-2+54)/4` `or` `x=(-2-54)/4`

`=>x=52/4` `or` `x=(-56)/4`

`=>x=13` `or` `x=-14`

`x` won't be negative because the value must be a positive integer.

So, `x=13`

Then `x+1=14`

So the positive consecutive integers whose sum of squares is 365 are `13` `&` `14.`

`=>13^2+14^2=365`

`=>169+196=365`

`=>365=365`

Answer 2

The positive integers are `13 and 14`

Explanation:

Let the two integers be `x and (x+1)`

The sum of their squares must be `365`

`x^2 + (x+1)^2 = 365`

`x^2 +x^2 +2x+1 =365" "larr` make the quadratic equal to `0`

`2x^2 +2x-364=0" "larr div 2`

`x^2 +x-182 =0`

As we are looking for two integers, the expression will factorise.

Factors which differ by `1` are on either side of the square root.

`sqrt182 =13.49`

The likely factors are `13 and 14,` check: `" "13 xx 14 = 182`

`(x-13)(x+14)=0`

Solving gives `x = 13 or x= -14` (which we reject as negative.)

`x =13 and x+1 = 14`