Question

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

Answer 1

Answer:

11 square + 13 square =121+169=290

solve by this method...so roots will be✓11 and√13

Answer 2

SOLUTION:

$\sf \: 11,13$

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.

$\sf \: ⇒ x² + (x + 2)² = 290$

$\sf \: → x² + x² + 4x + 4 = 290$

$\sf \: → 2x² + 4x + 4 - 290 = 0$

$\sf→ 2x² + 4x - 286 = 0$

Dividing by 2 we get,

$\sf \: → x² + 2x - 143 = 0$

Apply Quadratic Equation

$\sf⇒ x² + 13x - 11x - 143 = 0$

$\sf \: ⇒ x (x + 13) -11 (x + 13 ) = 0$

$\sf \: → (x-11) (x + 13 ) = 0$

$\sf \: x = 11 \: ,x + 2 = 11 + 2 = 13$

Topic:- Number System,Quadratic Equations.