# An easier way of solving a problem about the number of roots of a square equation on an interval

I am trying to solve the following problem: I need to find out the number of roots on the interval $[-1; 3)$ of the equation $(4-a)x^2-6ax+3=0$ depending from $a$. I know a solution, but it is too difficult to calculate. I thought I could solve that problem with the help of quadratic trinomial roots position theorem. In the end, I got a lot of inequalities and I got confused. Is there an easier way to solve this problem?

The quadratic formula yields $$x=\frac{6a\pm\sqrt{36a^2-12(4-a)}}{2(4-a)}.$$ Let us set this as two functions of $a:$ $$f_{\pm}(a)=\frac{6a\pm\sqrt{36a^2-12(4-a)}}{2(4-a)}.$$ These functions are undefined for $a=4.$ To find the exact domain, we factor what is under the square root: $$36a^2+12a-48=12(a-1)(3a+4).$$ This is a parabola opening up, and therefore it is negative in-between the two roots of $-4/3$ and $1$. So the domain of $f_{\pm}$ is $(-\infty,-4/3]\cup[1,4)\cup(4,\infty).$
It remains to find the range of $f_{\pm},$ which is what we really care about. This will be different depending on whether we are looking at $f_{+}$ or $f_{-}.$ For $f_{+},$ a simple plot reveals that $f_{\pm}(-4/3)=-3/4,\; f_{\pm}(a)=1,$ that $$\lim_{a\to-\infty}f_{+}(a)=0, \quad \lim_{a\to-\infty}f_{-}(a)=-6.$$ We can piece together the graphs of these functions on the range $[-1,3),$ draw vertical lines, and see how many times we intersect a function. From this we gather that for $a\in(-\infty,-7/5),$ there is only one root. For $a\in[-7/5,-4/3),$ there are two. For $a=-4/3,$ there is only one. For $a\in(-4/3,1),$ there are none. For $a=1,$ there is one. For $a\in(1,13/9],$ there are two. And finally, for $a\in(13/9,\infty),$ there is only one.
• @student28: The range of $f_{\pm}$ tells us where $x$ lies, because I defined $x=f_{\pm}(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting. – Adrian Keister Jul 18 '18 at 20:09
Try using the quadratic formula. This will give $x = \frac{3a \pm \sqrt{9a^2+3a-12}}{4-a}$.
Now you want to find values of $x \in [-1,3)$. So try setting $x=-1$ and $x=3$ and solve for $a$.
You can get a bound on $a$ because you are looking for real roots, and the discriminant must be nonnegative ($9a^2+3a-12 \geq 0$)
Using the above, you should be able to determine a range of values for $a$.