7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I find the hint to "complete the square" interesting; it seems to me unnecessary, as multiplying by four and using the given inequality suffices, and potentially confusing for readers unfamiliar with such a method. I have posted an answer to this effect. $\endgroup$ – Benjamin Dickman Mar 24 '18 at 23:57
13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That's a great point, and makes the follow-up statement very obvious, as opposed to my attempted proof which doesn't naturally suggest a basis for it. Thank you. $\endgroup$ – user242007 Mar 24 '18 at 20:11
  • $\begingroup$ Presumably, this approach could also be applied to the proof that $x^2 + xy + y^2 \gt 0$ for all $x \neq 0$, such as by rewriting the quadratic as $(x + \frac{y}{2})^2 + \frac{3y^2}{4}$, both terms necessarily being positive? $\endgroup$ – user242007 Mar 24 '18 at 20:17
  • 1
    $\begingroup$ @user242007 It's not the case that both terms there are always necessarily strictly positive. This fails to happen if $y = 0$ or $y = -2x$. However, since $x \neq 0$, in either of these cases we see that one of the terms is strictly positive whilst the other is non-negative so yes, the same approach works (up to a minor amendment). Of course, for $y \neq 0$ (which is the trivial case), this follows immediately from the above by taking $b = y, c = y^2$. $\endgroup$ – Rhys Steele Mar 24 '18 at 20:24
5
$\begingroup$

Given that $b^2-4c<0$,

$$x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\frac{b^2}{4}+c\ge-\frac{b^2}{4}+c=-\frac{1}{4}\left(b^2-4c\right)>0$$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

What you are missing is that$$x^2+bx+c=\left(x+\frac b2\right)^2-\frac{b^2-4c}4.$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ In this case using MathJax to format your question only involved adding $ signs around the things that should be in math mode. I've now done this and you can click on the edit button to see what you should have typed to format things. You should also read the link here to learn the basics of MathJax. As an aside, the fact that $b^2 < 4c$ means the quadratic only has imaginary solutions is already a less elementary fact than the question itself so it's probably desirable to avoid using it. $\endgroup$ – Rhys Steele Mar 26 '18 at 17:23
0
$\begingroup$

$$b^2-4c<0\le(2x+b)^2$$ $$\implies0<(2x+b)^2-b^2+4c=4x^2+4bx+4c$$ $$\implies x^2+bx+c>0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.