For what values of k does 3x^2- kx + 2 have two equal real roots?

[her]

I'm a junior in high school. I'm taking Pre-Calculus 11 online and I'm struggling with this question.

For what values of k does 3x²- kx + 2 have two equal real roots?

I've started using b² - 4ac, but I don't know where to go from there or if that's even the best way to approach this question.
For 3x² - kx + 2, a=3, b=k, c=2.
I know that for a quadratic equation to have 2 equal roots, b² - 4ac = 0.
k² - 4(3)(2) = b² - 4ac
k² - 4(3)(2) = 0
k² -24 = 0
k² = 24
k = √24

I don't know how to continue on from there. I would appreciate all the help I can get!

 

[mmm4444bot]

… For what values of k does 3x²- kx + 2 have two equal real roots?

… I know that for a quadratic equation to have 2 equal roots, b² - 4ac = 0

… k = √24

I don't know how to continue on from there …

That's not a correct expression for k.

\(\displaystyle \sqrt{k^2} = |k| \quad \text{not }k\)

Fix the sign issue. Then, do you know how to simplify the radical? (Most instructors would like to see the answers simplified.)

 

[mmm4444bot]

… a=3, b=k, c=2

… k² - 4(3)(2) = b² - 4ac

b = -k

So we should write b^2 as (-k)^2

Because -k is squared, in this exercise, you came out okay. Be mindful of coefficient signs. :cool:

 

[Dr.Peterson]

I'm a junior in high school. I'm taking Pre-Calculus 11 online and I'm struggling with this question.

For what values of k does 3x²- kx + 2 have two equal real roots?

I've started using b² - 4ac, but I don't know where to go from there or if that's even the best way to approach this question.
For 3x² - kx + 2, a=3, b=k, c=2.
I know that for a quadratic equation to have 2 equal roots, b² - 4ac = 0.
k² - 4(3)(2) = b² - 4ac
k² - 4(3)(2) = 0
k² -24 = 0
k² = 24
k = √24

I don't know how to continue on from there. I would appreciate all the help I can get!

Don't forget that there are two numbers whose square is 24!

Find both, and simplify the radicals. Otherwise, you're doing fine. I'd do just what you did.