What is the value of $E$ so that $4x^2-4x+E = 0$ has distinct real roots?

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Answer 1 · 2017-08-05T00:57:02

$E lt 1$

We have a quadratic:

$4x^2-4x+E = 0$

We have two unique (real) solutions if and only if the discriminant of the quadratic is positive. i.e

$Delta = b^2-4ac gt 0$
$=> (-4)^2-4(4)(E) gt 0$
$:. 16-16E gt 0$
$:. 1-E gt 0$
$:. E lt 1$

We can confirm this graphically, as $E=1$ should present two coincidental solutions, and $E gt 1$ no solutions:

Case $E=1$ - One Repeated Solution
graph{4x^2-4x+1 [-3, 3, -2, 5]}

Case $E=1.1$ - No Solutions
graph{4x^2-4x+1.1 [-3, 3, -2, 5]}

Case $E=0.9$ - Two Solutions
graph{4x^2-4x+0.9 [-3, 3, -2, 5]}

What is the value of EE so that 4x^2-4x+E = 0 4x^2-4x+E = 0 has distinct real roots?