Question

Find two consecutive odd positive integers, sum of whose squares is 290 by using quadratic formula

Answer 1

Answer:

11 and 13

Step-by-step explanation:

Let the two consecutive odd positive integers be x and x + 2

Now it is given that the sum of the squares is 290.

⇒ x² + ( x + 2 )² = 290

⇒ x² + x² + 4x + 4 = 290

⇒ 2x² + 4x + 4 - 290 = 0

⇒ 2x² + 4x - 286 = 0

Dividing by 2 we get,

⇒ x² + 2x - 143 = 0

⇒ x² + 13x - 11x - 143 = 0

⇒ x ( x + 13 ) -11 ( x + 13 ) = 0

⇒ ( x - 11 ) ( x + 13 ) = 0

⇒ x = -13, 11

In the question we are given with positive integers.

Hence the number is 11 and the consecutive odd number is 13.

Hence these are the required numbers.

Answer 2

Hello User!

Find two consecutive odd positive integers, sum of whose squares is 290 by using quadratic formula.

Solution:-

Let one of the odd positive integer be K.

&,Another odd positive integer be K + 2.

Their Sum of Squares,

⇒ k² + (k + 2)²
⇒ k² + k² + 4k + 4
⇒ 2k² + 4k + 4

Given, Their sum of squares = 290

⇒ 2k² + 4k + 4 = 290
⇒ 2k² + 4k = 290 - 4
⇒ 2k² + 4k = 286
⇒ 2k² + 4k - 286 = 0
⇒ 2(k² + 2k - 143) = 0
⇒ k²+ 2k - 143 = 0
⇒ k² + 13k - 11k - 143 = 0
⇒ k(k + 13) - 11(k + 13) = 0
⇒ (k - 11) = 0 And, (k + 13) = 0

Hence, K = 11 or -13

We always take positive value of K.

So, K = 11 And, (K + 2) = 11 + 2 = 13

Hence, The two consecutive positive integers are 11 and 13 .

Hope It Helps!