Question

The roots of each of the following quadratic equation are real and equal ,find k 3y^2+ky+12=0

Answer 1

$given - - > \\ \\ p(x) = 3 {y}^{2} + ky + 12 = 0 \\ \\ a = 3 \\ b = k \\ c = 12 \\ \\ it \: is \: given \: that \: roots \: of \: p(x) \: are \: real \: and \: equal \: \\ \\ so \\ = > {b}^{2} - 4ac = 0 \\ \\ = > {(k)}^{2} - 4(3)(12) = 0 \\ \\ = > {k}^{2} = 144 \\ \\ = > k = 12 \: \: \: (ans)$

Answer 2

Heya !!!



P(X) = 3Y² + KY + 12 = 0

Here,



A = 3 , B = K and C = 12


The roots of the given equation is real and equal it means Discriminant of the equation is greater than 0.



D = (B)² - 4AC



D = (K)² - 4 × 3 × 12




D = (K² - 144 )


Now,



(K²-144) = 0



K² = 144


K = ✓144


K = -12 OR 12.



HOPE IT WILL HELP YOU...... :-)