Question

Find two consecutive odd positive integers, sum of whose squares is 290. in thick can we take one no. x and other x+1

Answer 1

normally we know ,
odd number in the form of (2n +1)
where n is natural number .
so,
two consecutive odd numbers are (2n +1) and (2n +3)

A/C to Question,
(2n +1)² + (2n +3)² = 290

4n² + 4n +1 + 4n² + 12n + 9 = 290

8n² + 16n -280 = 0

n² + 2n - 35 = 0

n² + 7n -5n -35 = 0

n( n + 7) -5(n + 7) = 0

(n + 7)(n -5) = 0

n = 5 , and -7
but n is natural number and natural number can't be negative .
hence, n = 5

so, two consecutive odd numbers
(2n +1) = (2×5 + 1) = 11
(2n + 3) = (2 × 5 + 3) = 13

[ you can take x and x + 1 but where x is an odd number then it's possible ,
e.g x = 2n +1 ]

Answer 2

we can't take the other no. as x+1
....we have take other no. as x+2 because we have to find two consecutive odd no. ( they have even no. in between)



1st no. =x
2nd no. =x+2


x^2+(x+2)^2=290

x^2+(x^2+2x*2+2^2)=290[by (a+b)^2=a^2+2ab
+b^2]

2x^2+4x+4=290

2(x^2+2x+2)=290

x^2+2x+2= 145

x^2+2x+1+1=145

x^2+2x+1= 144

(x+1)^2= 144[by (a+b)^2=a^2+2ab +b^2]

x+1=√144

x+1=12

x=11


1st no.=11
2nd no.=13