Question

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

Answer 1

Answer:

11 & 13

Step-by-step explanation:

• let the two consecutive odd numbers be x, (x+2)
• sum of squares of those numbers is 290

${x}^{2} + {(x + 2)}^{2} = 290 \\ {x}^{2} + ( {x}^{2} + {2}^{2} + 2 \times x \times 2) = 290 \\ {x}^{2} + {x}^{2} + 4 + 4x = 290 \\ 2 {x}^{2} + 4x + 4 = 290$

divide the equation with 2

$2( {x}^{2} + 2x + 2) = 290 \\ {x}^{2} + 2x + 2 = \frac{290}{2} \\ {x}^{2} + 2x + 2 = 145 \\ {x}^{2} + 2x + 2 - 145 = 0 \\ {x}^{2} + 2x - 143 = 0$

${x}^{2} + 13x - 11x - 143 = 0 \\ x(x + 13) - 11(x + 13) = 0 \\ (x + 13)(x - 11) = 0 \\ x = - 13 \: \: \: \: \: {or} \: \: \: \: x = 11$

so the two consecutive odd numbers are 11 and 11+2= 13

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let x be an odd positive integer.

Then,

An odd positive integer just greater than x is x + 2.

According to the Question,

⇒ x² + (x + 2)² = 290

⇒ 2x² + 4x + 4 = 290

⇒ 2x² + 4x - 286 = 0

Dividing equation by 2, we get

⇒ x² + 2x - 143 = 0

By using factorization method, we get

⇒ x² + 13x - 11x - 143 = 0

⇒ x(x + 13) - 11(x + 13) = 0

⇒ (x + 13) (x - 11) = 0

⇒ x + 13 = 0 or x - 11 = 0

⇒ x = - 13, 11 (As x can't be negative)

⇒ x = 11

1st number = x = 11

2nd number = x + 2 = 11 + 2 = 13

Hence, the two consecutive numbers are 11 and 13.