Question

em (3K/2)x ² + (k-3)x - 5/2 = 0, k≠ 0 qual é o valor de k para que - 1/2 seja a raiz?

Answer 1

Temos o seguinte:

$\left ( \dfrac{3k}{2} \right)x^2 + (k-3)x - \dfrac{5}{2} = 0$

Substituindo a raiz, encontramos o valor de k:

$\left ( \dfrac{3k}{2} \right)* \left ( -\dfrac{1}{2} \right )^2 + (k-3)* \left ( -\dfrac{1}{2} \right ) - \dfrac{5}{2} = 0 \\ \\ \\ \left ( \dfrac{3k}{2} \right)* \left ( \dfrac{1}{4} \right ) + (k-3)* \left ( -\dfrac{1}{2} \right ) - \dfrac{5}{2} = 0 \\ \\ \\ \left ( \dfrac{3k}{8} \right) + (k-3)* \left ( -\dfrac{1}{2} \right ) - \dfrac{5}{2} = 0 \\ \\ \\ \left ( \dfrac{3k}{8} \right) + \left (\dfrac{-k + 3}{2} \right) = \dfrac{5}{2} \\ \\ \\ 3k - 4k + 12 = 20$

Logo:

$3k - 4k + 12 = 20 \\ \\ -k = 8 \\ \\ k = -8$