quadratic equation question

Firstly, you should have more working when you complete the square.

You are asked to prove that if $b^2<4ac$ then $f(x) >0$. What you have attempted to do instead is prove that if $f(x) >0$ then $b^2<4ac$ and then use this to prove $f(x) >0$.

But you don't need the first part since you're told already that $b^2<4ac$. The major problem is that knowhere in your working have you actually proved why $b^2<4ac$ implies that $f(x)>0$.

You could talk about the discriminant/no real roots to show this but I assume from your working that you need to prove it by completing the square? You correctly completed the squre but didn't use it for anything.

Think about why this

$\displaystyle =a\left ( x+\frac{b}{2a}\right )^2 + \frac{4ac-b^2}{4a}$

must always be positive.


One more thing:

$\sqrt{ -g} \rightarrow \mathbb{C}$

This doesn't mean much. I assume you're trying to say that $\sqrt{ -g}$ is a complex number but the set of complex numbers includes the reals and the imaginary numbers. What you mean to say is that $\sqrt{ -g}$ is imaginary. There is a mathematical way to write this but you may as well just write it in words and I recommend this unless you're confident with using the formal notation.

thank you very much for helping me
here goes my second attempt

I'm going to try explaining this to you because I don't feel like subtle hints are working very welly. (sorry for butting in, notnek!)

You're over complicating it. From

you know that

(i) the squared part is always greater than or equal to 0, because that's a basic property of a square.

(ii) you're multiplying the square by $a$, which you know is $>0$. So .

Now, it would be really nice if we could prove that $\frac{4ac - b^2}{4a}$ was also positive, in which case, we would know that $f(x)$ was a sum of two positive terms and hence would be positive itself.

But, how to do that... well, for a fraction to be positive, we want both its numerator and denominator to be positive. You can see trivially that the denominator is always positive here (because you're given that$a > 0$, so $4a$ must also be $> 0$).

You're reduced the entire problem to proving that the numerator is also positive. So, how can we prove that $4ac - b^2 > 0$? Oh... what a coincidence! This is precisely what we were assuming!

In writing the actual proof up, it would go something like this:

$f(x) = a\left ( x+\frac{b}{2a}\right )^2 + \frac{4ac-b^2}{4a}$, since $a > 0$ the first term is non-negative. Since $b^2 < 4ac$ we can then conclude that $4ac - b^2 > 0$ and so $\frac{4ac-b^2}{a} > 0$ also. So $f(x)$ is the sum of two positive terms and is hence positive itself.

thank you very much for helping me
here goes my second attempt

thank you woow i dont know what to say but thanks

can i conclude from what you said that if a quadratic function is a sum of two squares therefore f(x) > 0 is positive for all real values of x and thus will have no real roots

I am just trying to make note on this question

...

Notnek deserves it more than I do.



Indeed! (although this is not the case in this particular question) - here it's "if a function is a sum of two positive things for all real values of x, then the function is positive for all real values of x".

In another question, where you could have $f(x) = (x-a)^2 + (x-b)^2$ then you can't quite say that $f(x) > 0$, the best you can say is that $f(x) \geq 0$, because $y^2 \geq 0$, not $y > 0$.

However, if you had something like $f(x) = (x-a)^2 + (x-b)^2 + 1$ then you can definitely say that $f(x) > 0$ and so it has no real roots.

In fact, you use this technique a lot when you're asked to do stuff like "prove that $f(x) = x^2 + 2x + 2$ has no real roots". You do: $f(x) = x^2 + 2x + 1 +1 = \underbrace{(x+1)^2}_{\geq 0} + 1$ so that $f(x) \geq 0 + 1 > 0$ means that $f(x)$ has no real roots.

thank i gave him a rep so did u and RDK games

i am going to work harder i feel i got played around by bostock and chandler (making easy stuff look difficult)