(Q7)Find two consecuctive positive odd integers, sum of whose squares is 290.

Answer:

Let the first positive integer be x.Then, the second integer will be x + 2. According to the question,

x2(x + 2)2 = 290

x2 + x2 + 4x + 4 = 290

2x2 + 4x - 286 = 0

x2 + 2x - 143 = 0

which is a quadratic equation in x.

Using the quadratic formula x =
-b ± (b2 - 4ac)
2a
we get, x =
-2 ± (4 + 572)
2a
=
-2±576
2a
=
-2 ±
2a

x = or x= -13

But x is given to be positive odd integer.Therefore,x≠ -13

Thus, the two consecutive odd integers are and (x+2)=+2=.

Check:2+2 = 121 + 169 = 290.