**Q) Find two consecutive odd positive integers, the sum of whose squares is by using the quadratic formula.**

**Ans: **

Let’s consider the 1st odd positive integer be X

Then the next consecutive odd positive integers will be X + 2

Now it is given, that the sum of the squares of these two numbers is 290

∴ X^{2} + (X + 2)^{2} = 290

∴ X^{2} + (X^{2} + 4 X + 4) = 290

∴ 2 X^{2} + 4 X + 4 = 290

∴ 2 X^{2} + 4 X – 286 = 0

∴ X^{2} + 2 X – 143 = 0

∴ (X + 13) (X – 11) = 0

∴ X = – 13 and X = 11

Since the integer required to be positive,

therefore X -13 and X = 11

And the next integer is X + 2 = 11 + 2 = 13

**Hence, the two consecutive positive odd integers are 11 & 13.**

**Please do press “Heart” button if you liked the solution.**