Question

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

find the value of ‘k’.​

Answer 1

Answer:

$for \: equal \: roots \: \\ d = 0 \\ {b}^{2} - 4ac = 0 \\ {k}^{2} - 4(3)(12) = 0 \\ {k}^{2} - 144 = 0 \\ k = \sqrt{144} = 12 \\ therefore \\ k = 12$

Answer 2

Answer:

k= 12 or k= -12

Step-by-step explanation:

3$y^2$ + ky + 12 = 0

a = 3, b = k, c = 12

$b^2$ - 4ac = 0                                  _ _ _ _ _( given )

∴ $k^2$ - 4(3)(12) = 0

∴ $k^2$ - (12 x 12) = 0

∴        $k^2$ - 144 = 0

∴                 $k^2$ = 144

∴                 $k$  = $\sqrt144$                 _ _ _ _ _( taking square roots of both sides)

∴                 $k$  = ±12

∴        $k$ = 12               OR               $k$ = -12