Question

Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.

Answer 1

Let $x$ be the positive odd number and the consecutive odd number be $(x+2)$.

It is given that the sum of the squares of two consecutive numbers is $130$, therefore,

$(x{)}^2+(x+2{)}^2=130\Rightarrow x^2+x^2+4+4x=130(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow 2x^2+4x+4-130=0\Rightarrow 2x^2+4x-126=0\Rightarrow 2(x^2+2x-63)=0\Rightarrow x^2+2x-63=0\Rightarrow x^2+9x-7x-63=0\Rightarrow x(x+9)-7(x+9)=0\Rightarrow (x-7)=0,(x+9)=0\Rightarrow x=7,x=-9$

Since it is given that $x$ is a positive odd number, thus $x=7$.

Now, $x+2=7+2=9$

Hence, the two consecutive positive odd numbers are $7$ and $9$.