Question

resolva:
$kx ^{2} - 4x + 1 = 0$





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Answer 1

Explicação passo-a-passo:

$\sf \Delta=(-4)^2-4\cdot k\cdot1$

$\sf \Delta=16-4k$

$\sf \Delta=4\cdot(4-k)$

$\sf x=\dfrac{-(-4)\pm\sqrt{4\cdot(4-k)}}{2\cdot k}=\dfrac{4\pm2\sqrt{4-k}}{2k}$

$\sf x'=\dfrac{4+2\sqrt{4-k}}{2k}~\rightarrow~\red{x'=\dfrac{2+\sqrt{4-k}}{k}}$

$\sf x"=\dfrac{4-2\sqrt{4-k}}{2k}~\rightarrow~\red{x"=\dfrac{2-\sqrt{4-k}}{k}}$