Respuesta :
Answer:
9 and 11
Step-by-step explanation:
The sum of the squares of two consecutive positive odd integers is 202.
Let first number be x and other number is (x+2).
According to question,
[tex]x^2+(x+2)^2=202\\\\x^2+x^2+4+4x=202\\\\2x^2+4x+4=202\\\\2x^2+4x-198=0[/tex]
It is a quadratic equation. The solution of a quadratic equation is given by :
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
We have, a = 2, b = 4 and c = -198
So,
[tex]x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a}, \dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-4+ \sqrt{4^2-4\times 2\times (-198)} }{2(2)}, \dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\x=9, -11[/tex]
So, first positive odd integer is 9 and second one is (9+2)=11
