Less than four distinct roots

The polynomial $f_c(x)$ will fail to have four distinct complex zeros when the quadratic polynomial $g_c(x) \; = \; x^2 - (c^2-7c+11)x + (18-21c+8c^2-c^3)$ either has repeated roots or has $0$ as a root.

1. One of the roots of $g_c(x)$ is zero precisely when $c^3 - 8c^2 + 21c - 18 \; =\; (c-2)(c-3)^2 \;= \; 0$ and so precisely when $c=2,3$ .

2. The roots of $g_c(x)$ are repeated when its discriminant is zero, so that $\begin{array}{rcl} (c^2 - 7c + 11)^2 - 4(18 - 21c + 8c^2 - c^3) &=& 0 \\ c^4 - 10c^3 + 39c^2 - 70c + 49 &=& 0 \\ (c^2 - 5c+ 7)^2 &=& 0 \end{array}$ and this happens for two distinct values of $c$ which sum to $5$ .

Thus the sum of the possible values of $c$ is $2+3+5=10$ .

We desire to write it as (x^2-a)(x^2-b), to find out something about the roots. Expanding, we find that a+b=c^2-7c+11 and ab=18-21c+8c^2-c^3=(2-c)(c-3)^2. Since (2-c) and (c-3)^2 sum to c^2-7c+11, we have found our a and b. Now, the roots of the equation are, in terms of c, plus or minus square root of 2-c and plus or minus c-3. We have 3 cases. Either 2-c=c-2, so c=2. Or c-3=3-c, so c=3. Or root 2-c=c-3, which results in the quadratic c^2-5c+7=0. We take both roots, since the other just corresponds to if we had paired up root 2-c and 3-c. Their sum is 5 by vieta's, so our answer is 2+3+5=10.

Let $t=x^2$ , and consider $f_c(t) = t^2-(c^2-7c+11)t+(18-21c+8c^2-c^3)$

For $f_c(x)$ to have strictly less than four distinct complex zeroes, $f_c(t)$ must either have repeated roots, or have a root $t=0$ (because $0$ is the only complex number to make $x^2=t$ have repeated roots).

In the first case, $f_c(t)$ have repeated roots, so the discriminant must be $0$ . We will have $(c^2-7c+11)^2 - 4(18-21+8c^2-c^3)=0$ , or, after simplification, $c^4-10c^3+39c^2-70c+49=0$ . Observing that $49 = 7^2$ and that $70= 2\times 7\times5, 10=2\times 1\times 5$ , we can rewrite the equation as $(c^2-5c+7)^2=0$ , giving us $c^2-5c+7=0$ . Thus, $c=\frac{5 \pm \sqrt{-3}}{2} = \frac{5 \pm i\sqrt{3}}{2}$

In the second case, we have a root $t=0$ , so substituting this into $f_c(t)$ gives us $18-21c+8c^2-c^3=0$ . Factorizing, we get $(2-c)(c-3)^2 = 0$ , giving us $c=2$ or $c=3$ .

Therefore the sum of all the values of $c$ is $\frac{5 + i\sqrt{3}}{2} + \frac{5 - i\sqrt{3}}{2} +3 +2 =5+3+2=10$

For convenience write the given equation as : $x^4 -bx^2 +q$ where $b=c^2-7c+11$ and $q=18-21c+8c^2-c^3$

This implies that $x^2=\frac{b \pm \sqrt{b^2-4q} }{2}$

Now, we must have that either $b^2- 4q=0$ or $b=\pm \sqrt{b^2-4q}$ .

This is because if the discriminant is greater that, less than , but not equal to zero, then we will have $4$ distinct solutions to $x$ for all such values of $c$ . This is because the solutions for $x$ are as follows: $\left\{ (\frac{b+\sqrt{b^2-4q} }{2}), (\frac{(b - \sqrt{b^2-4q} }{2}),(\frac{-b+\sqrt{b^2-4q}}{2}), (\frac{-b-\sqrt{b^2-4q} }{2}) \right\}$ . Clearly these solutions will be distinct if the two condition mentioned before are not satisfied.

Case 1: $b^2-4q=0$

Note that a little algebraic manipulation gives us that $b^2-4q=(c^2-5c+7)^2$ .

Clearly, this equation has its discriminant less than zero and there are two different complex solutions to it whose sum by the vieta's formula are $5$ .

Case 2:

$b= \pm \sqrt{(b^2-4q)}$ $c^2-7c+11=\pm(c^2-5c+7)$

When $c^2-7c+11=c^2-5c+7 ,$ then $c=2$

When $c^2-7c+11=-(c^2-5c+7),$ then $2(c^2 - 6c +9)=(2(c-3)^2)=0$

Thus, since we are only interested in distinct solutions, $c=3$ .

Hence sum of all possible $c=5+2+3=\boxed{10}$

If we replace $x^2$ with $a$ , then it becomes quadratic with respect to $a$ . Since $a = x^2$ , then the two solutions of $a$ yield both a positive and a negative answer. Thus, to make $f_c(x)$ have less than four distinct complex zeroes, then either one of the roots of $f(a)$ must be $0$ or $f(a)$ has a double root.

Case 1 : If $0$ is a root, then, $-c^3 + 8c^2 - 21c + 18 = 0$ . Using Rational Root Theorem to factor this, we get $-(c - 3)^2(c - 2)$ as roots, thus $c = 3$ and $c = 2$ .

Case 2 : If $f(a)$ has a double root, then $(c^2 - 7c + 11)^2 + 4(c^3 - 8c^2 + 21c - 18) = 0 = c^4 - 10c^3 + 39c^2 - 70c + 49$ by the quadratic formula. By Vieta's formulas, the sum of the roots of this polynomial is $\frac{10}{2} = 5$ . Thus, the sum of the values of $c$ in this case are $5$ .

Alternative method If you're crazy, with a little bit of ingenuity, this can be factored to $(c^2 - 5c + 7)^2$ . Using the quadratic formula on this polynomial we get the roots to be $\frac{1}{2} (5 \pm i \sqrt{3})$ . Thus, the sum of these two roots is $5$ as found before.

Adding together the values of $c$ we found, we get $S_c = 2 + 3 + 5 = 10$ .

We note that the given expression is a quadratic in $x^2$ . We have $(x^2)^2-(c^2-7c+11)(x^2)+(18-21c+8c^2-c^3)$ We use the quadratic formula to solve this. $x^2=\dfrac{c^2-7c+11\pm\sqrt{(c^2-7c+11)^2-4(18-21c+8c^2-c^3)}}{2}$ Now, we expand the expression in the radical. I'll leave that part out; it's just algebra. If you want to determine that the expression is a square, you could plug in integers, and notice all outputs are squares. We get that $x^2=\dfrac{c^2-7c+11\pm (c^2-5c+7)}{2}$ We check each case separately.

Case 1: Evaluate as a plus sign.

Here, we have that $x^2=\dfrac{c^2-7c+11+c^2-5c+7}{2}=c^2-6c+9=(c-3)^2$ Hence, $x=\pm(c-3)$

Case 2: Evaluate as a minus sign.

Here, we have that $x^2=\dfrac{c^2-7c+11-c^2+5c-7}{2}=2-c$ Hence, $x=\pm\sqrt{2-c}$

Now, we have four solutions for $x$ in terms of $c$ . $\begin{aligned} &(1) x=c-3 \\ &(2) x=-(c-3) \\ &(3) x=\sqrt{2-c} \\ &(4) x=-\sqrt{2-c} \end{aligned}$ Now, our solutions are two pairs of additive inverses. Therefore, if all four are equal, they must all be 0. Clearly, this is not achievable. If three are equal, those three must be 0. Again, this is not achievable. Therefore, 2 must be equal.

Say solutions 1 and 2 are equal. Therefore, $c=3$ . This is one possibility.

Say solutions 3 and 4 are equal. Therefore, $c=2$ . This is another possibility.

Our other option would be if solution 1 equaled one of {3,4}, and therefore solution 2 would equal the other of {3,4}. $c-3=\pm\sqrt{2-c}$ Squaring this equation will save time, since it will solve both solution 1=3 and 1=4 at the same time. $c^2-6c+9=2-c$ $c^2-5c+7=0$ This has two complex solutions which sum to $5$ . Both are possible values of $c$ .

Therefore, the answer is $2+3+5=\boxed{10}$ .

The given function is a quadratic equation if we let x^2=y

This quadratic equation's roots can have these possible cases:

CASE 1: The quadratic equation has two distinct real roots or two distinct complex roots, which will be the values of y.

Since x=+-sqrt(y), each value of y will give two values of x, and so there will be a total of four different values of x, which we don't want.

CASE 2: Now, the only case that is left is that if the quadratic equation in y has equal roots. then x=+-sqrt(y) will return two distinct values only. This is what we want.

So, we apply the condition for equal roots.

(c^2-7c+11)^2=4(18-21c+8c^2-c^3)

This is a bi-quadratic equation. The sum of all roots of c will be given by:

-(coefficient of x^3 term)/(coeffecient of x^4 term) which is 10.