Question

use the discrminant to determine all values of k which would result in the equation -3x^2-6x+k=0 having real,unequal roots.​ must be right ty helppp asap

Answer 1

Answer:

$k>-3$

Step-by-step explanation:

We have the equation:

$-3x^2-6x+k=0$

Where a = -3, b = -6, and c = k.

And we want to determine values of k such that the equation will have real, unequal roots.

In order for a quadratic equation to have real, unequal roots, the discriminant must be a real number greater than 0. Therefore:

$b^2-4ac>0$

Substitute:

$(-6)^2-4(-3)(k)>0$

Simplify:

$36+12k>0$

Solve for k:

$12k > -36$

$k>-3$

So, for all k greater than -3, our quadratic equation will have two real, unequal roots.

Notes:

If k is equal to -3, then we have two equal roots.

And if k is less than -3, then we have two complex roots.