Question

If one root of 2x² + kx + 1 = 0 is -1/2 then the value of 'k' is
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Answer 1

$Given \: \frac{-1}{2}\: is\:a \: one \:root \:of \\2x^{2} + kx + 1 = 0$

$Substituting \: x = \frac{-1}{2} \:in \:the \\equation , we \:get$

$\implies 2\Big(\frac{-1}{2}\Big)^{2}+ k\Big(\frac{-1}{2}\Big)+1 = 0$

$\implies \frac{2}{4} - \frac{k}{2} + 1 = 0$

$\implies - \frac{k}{2} + 1+\frac{1}{2} = 0$

$\implies - \frac{k}{2} +\frac{2+1}{2} = 0$

$\implies - \frac{k}{2} +\frac{3}{2} = 0$

$\implies - \frac{k}{2} = -\frac{3}{2}$

$\implies k = \frac{-3}{2} \times \frac{-2}{1}$

$\implies k = 3$

Therefore.,

$\red { Value \: of \: k } \green {=3}$

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Answer 2

Answer:

If we substitute root in the polynomial we will get the zero.

=> 2 (-1/2)^2+k (-1/2)+1=0

=> 2 (1/4) + (-k/2)+1=0

=> 1/2 +1 -k/2 = 0

=> k/2 = 3/2

=> k = 3