Question 515532
So, for what value of c will {{{2x^2 - 3x + c = 0}}} have one and only one real root? To answer this question, let's think about the quadratic formula:

{{{x = (-b+- sqrt(b^2 - 4ac)) / 2a}}}

Specifically take note of the {{{sqrt(b^2 - 4ac)}}}. What's inside of this radical is known as the determinant, because it determines how many roots there will be.

{{{If}}} this {{{determinant}}} is {{{negative}}}, then there are {{{no}}}{{{ real}}} solutions. This is because you can't take the square root of a negative number without using imaginary numbers. Therefore, a negative determinant implies no real roots.

{{{If}}} this {{{determinant}}} is {{{positive}}}, then there are {{{2}}}{{{ real}}} solutions. This is because of the ± in the quadratic formula. 

{{{If}}} the {{{determinant}}} happened to be {{{1}}}, then you'd have {{{two}}}{{{ answers}}}: one you would have to add the {{{1}}}, and the other you would have to subtract the {{{1}}}.

{{{If}}} the {{{determinant}}} is {{{zero}}}, then there is {{{only}}}{{{ 1}}}{{{ real}}} solution. This is because + or - zero leaves the quadratic formula unchanged. In summary:

One real solution: {{{b^2 - 4ac=0}}}
Two real solutions:{{{ b^2 - 4ac>0}}}
No real solutions: {{{b^2 - 4ac<0}}}

So let's solve to see when the determinant equals zero, so there will be only one real root. We will use the a, b, and c values from the equation you provided earlier.

{{{2x^2 -3x + c = 0}}}
{{{a = 2}}}
{{{b = -3}}}
{{{c = Unknown}}}

Let's set the determinant equal to zero, and then substitute the known values:

{{{b^2 - 4ac = 0}}}
{{{(-3)^2 - 4(2)(c) = 0}}}
{{{(-3)^2 - 8c = 0}}}
{{{9 - 8c = 0}}}
{{{9 = 8c}}}
{{{9/8 = c}}}

{{{c}}} must be equal to {{{9/8}}} for the equation to have one and only one real root.