For what values of d will the roots of 1/4x^2 +dx-8 =0 be real?
Answer reads: All values

Not sure what I am doing wrong, i thought b^2-4ac 0 for real values...

Question

For what values of d will the roots of 1/4x^2 +dx-8 =0 be real?
Answer reads: All values

Not sure what I am doing wrong, i thought b^2-4ac >0 for real values...

anonymous · Answer

You're on the right track.
For this problem, your determinant is 
$$d ^{2}-4\times \frac{ 1 }{ 4 }\times(-8)$$
or
$$d ^{2}-8$$
So you're $$d ^{2}$$ must be greater than or equal to 8 in order to be positive and create real solutions.

marigirl · Answer

Hi, Thank you for taking the time out to answer my question.I substituted  in  values and got $$d ^{2}^{+8}\ge0$$

marigirl · Answer

$$d ^{2}+8\ge0$$

marigirl · Answer

I guess when you look at it...all  numbers will satithatsatisfy that inequation

directrix · Answer

I was looking at this problem because it is interesting.

I found the discriminant and got d² + 8.
A value of d that makes d² + 8 < 0 would yield a nonReal root.
 d² + 8 would never be negative or zero.
It appears that any d will do.

anonymous · Answer

Oops, I put minus instead of plus. Yep, you got it.

marigirl · Answer

That is why the answer reads All values,because all values will satisfy it. 
if we go on to solve it and find d, it becomes square root of a negative number ??? better to solve it by thinking about it?

directrix · Answer

I don't know your level of study but I think I can correctly say that the coefficients of the polynomials with which you will work for the time being will be real values. So, solving for d and getting a nonReal result would not be a value of d which applies to this problem. So, yes. I agree with you. Thinking about a problem conceptually is always appropriate and especially when interpreting an answer.