Find two consecutive odd positive integers, sum of whose squares is 290.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

11, 33

11, 53

11, 13

21, 13

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given two consecutive positive integers, sum of squares is 290.
Let one of the odd positive integers be x.
Then the other odd positive integer be

x + 2

.
The sum of square

= x^{2} + {(x + 2)}^{2}

\Rightarrow x^{2} + x^{2} + 4 x + 4

\Rightarrow 2 x^{2} + 4 x + 4

Given that their sum of squares

= 290

290 = 2 x^{2} + 4 x + 4

\Rightarrow 286 = 2 x^{2} + 4 x

\Rightarrow x^{2} + 2 x - 143 = 0

Now, factorising the equation,

\Rightarrow x^{2} + 13 x - 11 x - 143 = 0

\Rightarrow x (x + 13) - 11 (x + 13) = 0

\Rightarrow (x + 13) (x - 11) = 0

\Rightarrow (x + 13) = 0 or (x - 11) = 0

\Rightarrow x = - 13 or x = 11

We have to consider positive integer, so, x=11.
Thus, the other number is x+2=11+2=13.
So, value of odd integer is

11, 13

.
Therefore, option (3) is correct.

Watch 3-min video & get full concept clarity