Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
Find two consecutive positive integers, sum of whose squares is 365. Find two consecutive positive integers, sum of whose squares is 365.
Dear StudentLet us say, the two consecutive positive integers bexandx+ 1.Therefore, as per the given questions,x^2+ (x+ 1)^2= 365⇒x^2+x^2+ 1 + 2x= 365⇒2x^2+ 2x - 364 = 0⇒x^2+x- 182 = 0⇒x^2+ 14x- 13x- 182 = 0⇒x(x+ 14) -13(x+ 14) = 0⇒(x+ 14)(x- 13) = 0Thus, either,x+ 14 = 0 orx- 13 = 0,⇒x= - 14 orx= 13since, the integers are positive, soxcan be 13, only.x+ 1 = 13 + 1 = 14 two consecutive positive integers will be 13 and 14.Thanks
Let the first no. = XSecond no. = X+1 (consecutive number mean continuously)According to the question(X)2 +(X+1)2 = 365X2+X2+1+2X = 3652X2+2X=365-12X2+2X=364Dividing both sides by 2X2+X=182X2+X-182=0Now cwe factorize this equationX2+14X-13X-182=0X(X+14)-13(X+14)=0(x+14)(x-13)=0X+14=0x=-14x-13=0x=13Given in question positive integer so we select 13 which is value of x so can say thatX=13X+1=13+1=14 Thank you 2X2+2X+1=365
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -