4) If roots of a quadratic equation 3y2+ ky+ 12= 0 are real and equal then

find the value of ‘k’.​

2 thoughts on “4) If roots of a quadratic equation 3y2+ ky+ 12= 0 are real and equal then <br /><br />find the value of ‘k’.​”

1. Answer:

   [tex]for \: equal \: roots \: \\ d = 0 \\ {b}^{2} – 4ac = 0 \\ {k}^{2} – 4(3)(12) = 0 \\ {k}^{2} – 144 = 0 \\ k = \sqrt{144} = 12 \\ therefore \\ k = 12[/tex]

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2. Answer:

   k= 12 or k= -12

   Step-by-step explanation:

   3[tex]y^2[/tex] + ky + 12 = 0

   a = 3, b = k, c = 12

   [tex]b^2[/tex] – 4ac = 0                                  _ _ _ _ _( given )

   ∴ [tex]k^2\\[/tex] – 4(3)(12) = 0

   ∴ [tex]k^2[/tex] – (12 x 12) = 0

   ∴        [tex]k^2[/tex] – 144 = 0

   ∴                 [tex]k^2[/tex] = 144

   ∴                 [tex]k[/tex]  = [tex]\sqrt144[/tex]                 _ _ _ _ _( taking square roots of both sides)

   ∴                 [tex]k[/tex]  = ±12

   ∴        [tex]k[/tex] = 12               OR               [tex]k[/tex] = -12

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