Question

The sums of squares of two consecutive positive odd numbers is 290. Find the number..​

Answer 1

The sums of squares of two consecutive positive odd numbers is 290. Find the number.

$\huge\bold{\underline{Answer:}}$

$\green\bigstar$ Given:

The sums of squares of two consecutive positive odd numbers is 290

$\blue\bigstar$ To find:

Find the number.

$\purple\bigstar$ Solution:

Let the two consecutive number be x and (x+2)

According to question, we have

$\sf{:\implies x²+(x+2)²=290}$

$\sf{:\implies 2x²+4x+4=290}$

$\sf{:\implies 2x²+4x=290-4}$

$\sf{:\implies 2x²+4x=286}$

$\sf{:\implies 2(x²+2x)=286}$

$\sf{:\implies x²+2x=143}$

$\sf{:\implies x²+2x-143=0}$

$\sf{:\implies x²+13x-11x-143=0}$

$\sf{:\implies x(x+13)-11(x+13)=0}$

$\sf{:\implies (x+13)(x-11)=0}$

$\sf{:\implies x=-13}$

or,

$\boxed{\bf{\pink{⟹\:x=11}}}$

Since the number is positive,the number is x = 11

.°. x + 2 = 11 + 2 = 13

Therefore,

sum of numbers is = ( 11 + 13 ) = 24

Therefore, the number is 24

Answer 2

Answer:

Let one of the odd positive integer be x 

then the other odd positive integer is x+2

their sum of squares

=x2+(x+2)2

             

=x2+x2+4x+4                          

=2x2+4x+4

Given that their sum of squares = 290

2x2+4x+4=290 

2x2+4x=286

2x2+4x−286=0

x2+2x−143=0

x2+13x−11x−143=0

x(x+13)−11(x+13)=0

(x−11)=0,(x+13)=0

Therfore ,x=11or−13

We always take positive value of x 

So , x=11 and (x+2)=11+2=13

 

Therefore , the odd positive integers are 11 and 13