My teacher was explaining quadratics in my class and it was a little bit unclear to me. The problem was

Suppose $at^2 + 5t + 4 > 0$, show that $a > 25/16$ .

My teacher said that there are no solutions for this function when it is greater than $0$ and used $b^2-4ac \lt 0$, and this is the part that confused me. I understand why he used $b^2-4ac \lt 0$ but I cannot understand why there are no solutions by just looking at the function. Could someone explain this to me?

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Edit: I will try to say something general first. Hopefully that will make my answer a bit easier to understand. But as a first thing, it might be helpful to take a look at for example the Wikipedia article on quadratic equations.

If you have a function $f(t)$, then $t$ is a variable (called the independent variable). When we change the value of $t$, then the value of $f(t)$ changes. For example if $f(t) = t +2$, then for $t = 1, f(t) = f(1) = 3$. For $t = 2, f(t) = t(2) = 4$.

Now given a function $f(t)$, you can ask the question: does there exist a value or $t$ such that for that value $f(t)$ is zero. You are asking if there are any zeros. Anything that makes that function equal to zero.

A special type of function are the quadratic polynomials. Here $f(t)$ looks like this:

$$f(t) = at^2 + bt + c.$$ ($a,b,c$ are fixed constants.) For example $f(t) = t^2 + 3t + 2$. Now we ask the question: does there exist a number such that the value of $f(t)$ is $0$? The answer is yes because $$f(-1) = (-1)^2 + 3\cdot(-1) + 2 = 0.$$

Now if you know about a function that $f(t)>0$ for all $t$, then no matter what $t$ is $f(t)$ is never going to be equal to zero. It is always positive. That is exactly what it means to say that $f(t) > 0$ for all $t$.

Another way to answer the question about whether a function is ever equal to zero is to draw the graph of the function. If the function intersects the $t$-axis (think: $x$-axis) then there is some number for which the value of the function is zero. That is exactly what it means to intersect the $t$-axis.

So to echo what other people have already mentioned: you are considering a polynomial: $$P(t) = at^2 +5t + 4.$$ If you sketch the graph of this polynomial you get a parabola. The question can be asked whether the graph of the polynomial intersects the $t$-axis. Asking this question is the same as asking for roots (or zeros) of the polynomial. I.e. you want to know whether of not there is a number $t_0$ such that $P(t_0) = 0$.

You are assuming (supposing) that $P(t) > 0$ for all $t$. I.e. that means that there are no roots of the polynomial, i.e. no solution to the equation $P(t) =0$. First you might notice that if $P(t) >0$ for all $t$, then $a$ has to be a positive number. If $a$ was negative, then for $t = -100000$, $P(t)$ would be negative, right?

So what more can we say? We have the discriminant that tells you something about how many roots the polynomial has. For a general quadratic equation: $at^2 + bt + c= 0$ ($a,b,c$ being constants here and $t$ the variable), the discriminant is defined as $$d = b^2 - 4ac.$$

It is a fact that if \begin{align} d &> 0\quad \text{ means that there are exactly two distinct roots}\\ d &= 0\quad \text{ means that there are exactly one root}\\ d &< 0\quad \text{ means that there are no roots.} \end{align}

For your polynomial, you would then need $d = 5^2 - 4a\cdot 4 < 0$. So \begin{align} 5^2 - 4a\cdot 4 &< 0 &\Rightarrow\\ - 4a\cdot 4 &< -25 &\Rightarrow\\ a &> \frac{25}{16}. \end{align} And there you go.

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So, since the function is > 0, there are no solutions because the solutions for all quadratics are on the x axis? – Kot Sep 22 '12 at 23:08
@StevenN: Since the function (polynomial) is greater than 0 for all $t$, then you know that it is never equal to $0$. So there are no roots of the polynomial (no zeros of the function), so therefore the discriminant must be negative. Does that make sense? – Thomas Sep 22 '12 at 23:13
@StevenN: I think that i understand your comment better now. If you by "on the $x$-axis" mean that the roots/zeros are where the graph intersects the $x$-axis, then you are right. – Thomas Sep 22 '12 at 23:15
I am still confused, could you explain it in more simple terms? I had very bad math teachers in high school and they didn't do a good job explaining concepts. – Kot Sep 22 '12 at 23:18
@StevenN: I will try to edit my answer a bit, but it might be helpful to first take a look at the Wikipedia article on quadratic equations: en.wikipedia.org/wiki/Quadratic_equation. – Thomas Sep 22 '12 at 23:23

The question really means that $at^2+5t+4>0$ for all real values of $t$. This is only possible if $a>0$ and the quadratic curve does not intersect the horizontal axis. In this case there are no real solutions to the equation $at^2+5t+4=0$ so that the discriminant is negative.

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In order for $f(t)=at^2+5t+4>0$ all $t$, then the discriminant must be negative, else there are real roots to $f(t)$. The discriminant is $25-16a$, so $25-16a<0$ and thus $a>25/16$.

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The equation a$t^2$ + 5t + 4 = 0 may or may not have a solution. This depends on the value of a that you pick. Graphically, if this equation has a solution, the graph will touch and/or go through the x-axis. If there is not a solution, the graph will not have a solution. These x-intercepts are defined by the solutions: $$x = {-b \pm \sqrt{b^2-4ac}\over2a}$$ For these solutions to be real, $b^2-4ac$, the discriminant, must be greater than or equal to 0. If not, the solution will be irrational(sqrt of neg number).

Hope this helps!

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Do you know about LaTeX? See: meta.math.stackexchange.com/questions/5020/… – Thomas Sep 22 '12 at 22:49
Thanks @thomas , I've been looking for a good tutorial. – Ben Sep 25 '12 at 23:57