Question

Find two consecutive odd positive integers sum of whose square is 290

Answer 1

Step-by-step explanation:

Let the two consecutive odd numbers are a and a + 2

Given, sum of square of two consecutive odd numbers is 290

=> a2 + (a + 2)2 = 290

=> a2 + a2 + 4a + 4 = 290

=> 2a2 + 4a + 4 = 290

=> 2a2 + 4a + 4 - 290 = 0

=> 2a2 + 4a - 286 = 0

=> 2(a2 + 2a - 143) = 0

=> a2 + 2a - 143 = 0

=> a2 + 13a - 11a - 143 = 0

=> a(a + 13) - 11(a + 13) = 0

=> (a + 13)*(a - 11) = 0

=> a = 11, -13

Since number is positive,

So, a = 11

Now, a + 2 = 11 + 2 = 13

So, the numbers are 11 and 13

Answer 2

Appropriate Question:

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

Solution:

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

According To The Statement, $\bf{ {x}^{2} + {(x + 2)}^{2} = 290 }$

$: \implies \bf{ {x}^{2} + {x}^{2} + {2}^{2} + 2 \times x \times 2 = 290 } \\ : \implies \bf{ {2x}^{2} + 4 + 4x = 290} \\ : \implies \bf{ {2x}^{2} + 4x - 286 = 0} \\ : \implies \bf{2( {x}^{2} + 2x - 143) = 0} \\ : \implies \bf{ {x}^{2} + 2x - 143 = 0 } \\ : \implies \bf{ {x}^{2} + 13x - 11x - 143x = 0} \\ : \implies \bf{x(x + 13) - 11(x + 3) = 0} \\ : \implies \bf{(x - 11)(x + 13) = 0} \\ : \implies \bf{x - 11 = 0} \: \: \: \sf{or} \: \: \: \bf{x + 130} \\ : \implies \bf{x = 11} \: \: \: \sf{or} \: \: \: \bf{x = - 13 \bigg(Rejected\bigg)} \\ \bf{ \therefore \: x = 11}$

Hence, Required Odd Consecutive Integers are 11 and 13.