Find the values of \( k \) that make [tex]2x^{2} + kx + 7[/tex] factorable.

the values of k that make > 2x² + kx + 7 factorable the values of k are 9 & 15 thus (2x² + 9x + 7) = (2x + 7)(x+ 1) k -9 (2x² + 15x + 7) = (2x +1)(x+7) K =15

Answered by Poknoy | 2024-06-10

1 ) ( 2 x + 1 ) ( x + 7 ) = 2 x 2 + 14 x + x + 7 = 2 x 2 + 15 x + 7 ⇒ k = 15 2 ) ( 2 x − 1 ) ( x − 7 ) = 2 x 2 − 14 x − x + 7 = 2 x 2 − 15 x + 7 ⇒ k = − 15 3 ) ( 2 x + 7 ) ( x + 1 ) = 2 x 2 + 2 x + 7 x + 7 = 2 x 2 + 9 x + 7 ⇒ k = 9 4 ) ( 2 x − 7 ) ( x − 1 ) = 2 x 2 − 2 x − 7 x + 7 = 2 x 2 − 9 x + 7 ⇒ k = − 9 A n s . t h ere a re f o u r so l u t i o n s : k ∈ { 15 ; − 15 ; 9 ; − 9 }

Answered by kate200468 | 2024-06-10

The values of k that make the quadratic expression 2 x 2 + k x + 7 factorable are k = 15 , 9 , − 15 , − 9 . These values meet the requirement for the discriminant to be a perfect square. The discriminant is defined as D = k 2 − 56 , which must equal a non-negative integer squared for the quadratic to be factorable.
;

Answered by kate200468 | 2024-12-23