Question

@Mathematics Using the discriminant, what values of c will make the quadratic y = 2x2 - 3x + c have one real solution?
A)c = -9/8
B)c = 9/8
C)(9/8, ∞)
D)[9/8, ∞)

Answer 1

If it has one real solution, then the discriminant is zero. Remember that:\[\Delta = b^{2} - 4ac\]Substitute the values in, and you will get:\[9 - 8c = 0\]Solve for c :-)

Answer 2

I'm still not understanding... Can you explain more in detail?

Answer 3

Bah, site crashed on me. So, remember that the quadratic formula is:\[x = (-b \pm \sqrt{\Delta})/2a\]In order for this to have one solution the plus or minus part has to be 0, that is, \[\sqrt{\Delta} = 0 \rightarrow \Delta = 0\]Do you understand it so far?

Answer 4

Yes.

Answer 5

So, recall that\[\Delta = b^{2} - 4ac\]All quadratic equations have the form:\[ax^2 + bx + c\]From your quadratic, we get: a = 2, b = -3 and c = c.

Answer 6

Now, because the delta must be equal to zero, we get\[\Delta = 0 \rightarrow (-3)^{2} - 4(2)c = 0 \rightarrow 9 - 8c = 0\]and we can solve for c.

Answer 7

And so C would equal 9/8 ?

Answer 8

Yes

Answer 9

But did you understand it more clearly now? I suck a bit at explaining something step by step, because I skip some every once in a while.

Answer 10

Oh. Yes, it was fine... Question though is 9/8 positive?

Answer 11

Yeah, it is. The answer is B. It's like:\[-8c = -9 \rightarrow c = (-1/-1)(9/8) = 9/8\]