Calculating range of values of $k$ s.t. the graph $y=4x^2-kx+25$ does not cut or touch the $x$ axis

Question

Calculate the range of values of $k$ so that the graph $y=4x^2-kx+25$ does not cut or touch the $x$ axis.

I just don't know what to set delta to as I can't work out if the graph would be a tangent to the $x$ axis or cut it in two places.

Any help would be much appreciated as I want to understand this topic. Thanks.

Answer 1

Hint:

The points of the $x$ axis are points with the $y$ coordinate null. So, If the graph does not cut or touch the $x$ axis, than we have no value $x$ such that $$ 0=4x^2-kx+25 $$

and this means that the discriminant of this equation is negative. can you do from this?

Answer 2

The short way is noting that statement is equivalent to find the values of $k$ for which discriminant of $4x^2-kx+25=0$ is negative (that is, solve $k^2-4\times 4\times25<0$)

Another approach is observing that $f(x)=4x^2-kx+25$ is a parabola which opens versus above. Then, if the minimum is >0, done. But $4x^2-kx+25=4(x^2-\frac{kx}{4}+\frac{k^2}{64}+\frac{25}{4}-\frac{k^2}{64})=4((x-\frac{k}{8})^2+\frac{25}{4}-\frac{k^2}{64})\ge4(\frac{25}{4}-\frac{k^2}{64})$

That is, $\min_{x\in\mathbb{R}}f(x)=4(\frac{25}{4}-\frac{k^2}{64})$ and is obtained for $x=k/8$. Then, you will search for $k$ such that $4(\frac{25}{4}-\frac{k^2}{64})>0$, which is equivalent to $400-k^2>0$

Answer 3

You might want to use the discriminant to solve this. So we have $y=4x^2-kx+25$

If the graph does cut the x-axis, or is tangent to it, it will equal $0$ at some point, meaning $0=4x^2-kx+25$

Now we want the values of $k$ for which there are no real solutions to this equation, as if there are no real solutions for $x$, it would mean the graph doesn't cut the x-axis.

So by the discriminant, for $ax^2+bx+c$, there are no real solutions if $$b^2-4ac<0$$

This would be because from the quadratic formula, $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ We can see that if the discriminant is negative (less than $0$), the solution would be complex.

So substituting our values into the discriminant, in order for $x$ to have no solutions, we get

$$(-k)^2-4(4)(25)<0 \Rightarrow k^2-400<0 \Rightarrow (k+20)(k-20)<0$$

Therefore, the graph won't cut or touch the x axis when $-20<k<20$, which we get by solving the above inequality.