Find k

Find k for 114. Something tells me to use the radicand of the quadratic formula. Yes?

 

Didn't you just post this?

 

I posted the same attachment but not the same question. Read what I said.

 

"Find k so that x^2- kx+ 4= 0 has a repeated real solution."

A quadratic equation has a repeated solution if and only if it is of the form (x- a)^2= 0 for some a. What must "a" be for x^2- kx+ 4= 0? So what is k?

 

I think the discriminant is needed.
If b^2-4ac = 0, there will be two equal real number solutions.

Let me see.

a =1, b = -k, c = 4

b^2-4ac = (-k)^2 - 4*1*4 =k^2-16

k^2-16 = 0

(k-4)(k+4) = 0

True when k = 4 and when k = -4.
Conclusion: There are two equal real number solutions when
k = 4 and when k = -4.

 

It's not that complicated! (x- a)^2= x^2- 2ax+ a^2. Comparing that to x^2- kx+ 4, we see that a^2= 4 so a is either 2 or -2 and so k is either 4 or -4. Of course if k= -4, that would be x^2+ 4x+ 4 so they probably would not have written "-kx". I suspect k= 4 was intended.

 

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3. If you are going to help me with high school probability review in just a few months from now, please show humility in your reply. Do your best to understand that not everyone has the same level of comprehension.

4. This thread is easy for you but not for me.

5. I know you from another forum. I just forgot where I saw you. Freemathhelp.com???