Question

Valor do k para que a função x²-5x+k admite 1/4?.

Answer 1

Resposta:

k=19/16

Explicação passo a passo:

$x^2-5x+k=0\\\dfrac{1}{4}^2-5(\dfrac{1}{4})+k=0\\ \dfrac{1}{16} -\dfrac{5}{4} +k=0\\\dfrac{1-5\times4 +16k}{16} =\dfrac{0}{16} \\\\1-20 +16k=0\\\\16k=19\\\\k=\dfrac{19}{16}$