Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality,  $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
 A: Given that $b^2-4c<0$,
$$x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\ge-\frac{b^2}{4}+c=-\frac{1}{4}\left(b^2-4c\right)>0$$
A: From here: $(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$ we get $b^2-4ac \geq 0$ and not $>$. So if $b^2-4c<0$ there is no real solution. So else from that $>$ your conclusion is correct.
A: Suppose for the sake of contradiction that $x^2 + bx + c = 0$ for the given hypothesis of $4c > b^2$.
Multiplying the given equation by $4$ and making use of our hypothesis yields:
$$0 = 4x^2 + 4bx + 4c > 4x^2 + 4bx + b^2 = (2x + b)^2$$
But, we cannot have that $0$ is strictly greater than a real number squared. Contradiction. QED.
A: Essentially the same algebraic manipulations used in the second part of your question will give a proof without aiming for a contradiction. 
Indeed, these manipulations will give you that $x^2 + bx + c = (x + \frac{b}{2})^2 - \frac{b^2 - 4c}{4}$. This is strictly positive since $(x + \frac{b}{2})^2 \geq 0$ and $\frac{b^2 - 4c}{4} < 0$. In particular, $x^2 + bx + c > 0$ for every $x$ which completes the proof.
A: What you are missing is that$$x^2+bx+c=\left(x+\frac b2\right)^2-\frac{b^2-4c}4.$$
A: A completely informal approach:
consider $x^2+bx+c$, now differentiate with respect to $x$ twice. This gives a positive value,which indicates that the parabola is upturned. Now consider the fact that $b^2<4c$, clearly gives imaginary solutions. Now if the parabola doesn't touch the $X$ axis and is upturned it must lie above the $X$ axis thus proving the fact that $f(x)>0$,because an upturned parabola whose minima lies below the '$X$' axis will always intersect the $X$ axis.
A: $$b^2-4c<0\le(2x+b)^2$$
$$\implies0<(2x+b)^2-b^2+4c=4x^2+4bx+4c$$
$$\implies x^2+bx+c>0$$
