Question #852f1

2 Answers
Aug 12, 2017

1313 && 1414

Explanation:

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

x^2+(x+1)^2=365x2+(x+1)2=365

We know that (a+b)^2=a^2+2ab+b^2.(a+b)2=a2+2ab+b2.So the equation becomes

=>x^2+(x^2+2*x*1+1^2)=365x2+(x2+2x1+12)=365

=>x^2+x^2+2*x*1+1^2=365x2+x2+2x1+12=365

=>x^2+x^2+2x+1=365x2+x2+2x+1=365

=>2x^2+2x+1=3652x2+2x+1=365

=>2x^2+2x+1color(red)(-365)=365color(red)(-365)2x2+2x+1365=365365

=>2x^2+2x-364=02x2+2x364=0

Apply Quadratic Formula, that is x=(-b+-sqrt(b^2-4ac))/(2a).x=b±b24ac2a. Here

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

=>x=(-2+-sqrt(2^2-(4*2*(-364))))/(2*2)x=2±22(42(364))22

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

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

=>x=(-2+-sqrt(4+2912))/4x=2±4+29124

=>x=(-2+-sqrt(2916))/4x=2±29164

=>x=(-2+54)/4x=2+544 oror x=(-2-54)/4x=2544

=>x=52/4x=524 oror x=(-56)/4x=564

=>x=13x=13 oror x=-14x=14

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

So, x=13x=13

Then x+1=14x+1=14

So the positive consecutive integers whose sum of squares is 365 are 1313 && 14.14.

=>13^2+14^2=365132+142=365

=>169+196=365169+196=365

=>365=365365=365

Aug 12, 2017

The positive integers are 13 and 1413and14

Explanation:

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

The sum of their squares must be 365365

x^2 + (x+1)^2 = 365x2+(x+1)2=365

x^2 +x^2 +2x+1 =365" "larrx2+x2+2x+1=365 make the quadratic equal to 00

2x^2 +2x-364=0" "larr div 22x2+2x364=0 ÷2

x^2 +x-182 =0x2+x182=0

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

Factors which differ by 11 are on either side of the square root.

sqrt182 =13.49182=13.49

The likely factors are 13 and 14, 13and14, check: " "13 xx 14 = 182 13×14=182

(x-13)(x+14)=0(x13)(x+14)=0

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

x =13 and x+1 = 14x=13andx+1=14