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I have a function: $f(x)=-\frac{4x^{3}+4x^{2}+ax-18}{2x+3}$ which has only one point of intersection with the $x$-axis.

How can i find the value of $a$?

I tried polynomial division and discriminant, but it didn't help me.

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find the value of $a$ that does what? –  Git Gud Apr 10 '13 at 21:04
    
@GItGud that satisfies unique intersection with x-axis –  user64066 Apr 10 '13 at 21:05
    
You could in principle use the cubic formula. –  Javier Badia Apr 10 '13 at 21:05
    
sorry, I miscalculated... –  Bitwise Apr 10 '13 at 21:21
    
Does the question say whether there is only one value of $a$, or are you supposed to determine how many there are and find them? –  Javier Badia Apr 10 '13 at 21:22
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up vote 0 down vote accepted

This may be a little longer than needed, but I wanted to show my thought process in finding the solution.

First of all, let's ignore the minus sign, it doesn't make any difference. Call $P(x) = 4x^3+4x^2+ax-18$ and $Q(x) = 2x+3$. Note that if $x \neq -\frac32$, then $Q(x) \neq 0$ and so we can pretty much ignore it when looking for roots.

Suppose $x=-\frac32$. Then $P(-\frac32) = 0$ if and only if $a = -15$ (you can check this by direct evaluation). We can factor $P$ in this case: $P(x) = 4(x+\frac32)^2(x-2)$. $Q$ can also be written as $Q(x) = 2(x+\frac32)$, so we can cancel $x+\frac32$ and find that while the function is not defined at $x=-\frac32$, its limit is $0$. $f(x)$ is also zero at $x = 2$, so if you count $\frac32$ as an intercept then $a=-15$ isn't a solution, while if you count it then it is.

Now suppose $x\neq -\frac32$. The problem reduces to $P(x) = 4x^3+4x^2+ax-18 = 0$. This is cubic equation, so it's gonna be pretty hard to solve it directly. Instead, let's look at the derivative: $P'(x) = 12x^2+8x+a$. If we set that equal to zero we get $x = \frac{-2\pm\sqrt{4-3a}}{6}$. If $a \gt \frac43$, this has no roots and is always positive. Therefore $P$ is increasing and it has only one root.

If $a = \frac43$ then $P'$ has a double root. $P$ is still increasing (though not strictly so), and therefore has only one root.

It seems like if $-15 \lt a \lt \frac43$ there's only one root, and if $a \lt -15$ there's three. I don't know how to prove this yet, so I'm posting this incomplete answer hoping it will be helpful.

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How many roots are there if $a\geq\frac{4}{3}$? –  user64066 Apr 10 '13 at 22:13
    
It says right there: $P$ (and therefore $f$) has only one root in that case. –  Javier Badia Apr 10 '13 at 22:17
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I played around with the numerator of this function in Mathematica. Lots of values of $a$ give one root, but only one value of $a$, $a=-15$, gives two distinct real roots:

cubeplot

The double root is at $x=-3/2$, the single at $x=2$. I suspect this is what you were looking for. In this case,

$$f(x) = (2 x+3) (x-2)$$

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First of all thanks a lot for the effort. i wonder if there's any Way to get this result without Mathematica. –  user64370 Apr 10 '13 at 21:53
    
In hindsight, you want the polynomial to be zero when $x=-3/2$. So you could plug in that value and solve for $a$. –  Ron Gordon Apr 10 '13 at 21:55
    
The problem is that if $a \lt -15$ there are three roots, so those values don't work. –  Javier Badia Apr 10 '13 at 22:00
    
@JavierBadia: sorry, I don't understand what you are getting at. –  Ron Gordon Apr 10 '13 at 22:04
    
The question asks for what values of $a$ there's only one root. Your answer seems to imply (sorry if I misinterpreted!) that except for $a = -\frac32$, every other value will work. –  Javier Badia Apr 10 '13 at 22:08
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I think this might work. The horizontal asymptote of this function is at $x =\frac{-3}{2}$. So whenever this division yields $0$ remainder this asymptote would not work since this means that we have some form of $(2x+3)(\alpha x^{2}+\beta x+ \theta)$. If this is the case than asymptote would not work out. So by polynomial division when $a =-15$ there is no remainder, however as long as $a>-15$ there is remainder and asymptote will work.

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