Question

Please solve it fast.........

Answer 1

Given Quadratic equation is 3x^2 + 2kx + 27 = 0.

Here, a = 3, b = 2k, c = 27.

We know that for real and equal roots, we need to solve the value of k such that Δ = 0.

⇒ b^2 - 4ac = 0

⇒ (2k)^2 - 4(3)(27) = 0

⇒ 4k^2 - 324 = 0

⇒ k^2 - 81 = 0

⇒ k^2 = 81

⇒ k = +9,-9.

Hope it helps!

Answer 2

hey mate!

3x^2+2x+27
real and equal roots

D=0

D= underroot b^2-4ac

D= underroot (2k)^2-4ac

D= underroot 4k ^2 - 324

so,

underroot 4k^2 - 324 =0

4k^2 -324=0

k^2 =324/4

k^2= 81

k= +9 or k = -9

for real and equal roots.

# Phoenix