A wonky quartic

Let $f(x) = (x^2-x+a+1)^2-4 a (5 x^2-x+1)$

After rearranging the terms we can write it as,

$\begin{array}{cc} f(x) &= (x^2-x+1-a)^2-16 a x^2\\ &= (x^2-(1-4\sqrt{a})x+1-a)(x^2-(1+4\sqrt{a})x+1-a) \end{array}$

If equation $f(x)=0$ have 3 distinct real roots then it can't have any complex root so its coefficients must also be real as well as coefficients of its factors. So, $\sqrt{a}$ is real, which means $a \geq 0$ .

Let $a=b^2$ where $b \geq 0$ , now $f(x) = (x^2-(1-4b)x+1-b^2)(x^2-(1+4b)x+1-b^2)$ have 4 roots which are,

$\begin{array}{cc} x_1&=\frac{1-4b-\sqrt{(20b^2-8b-3)}}{2}\\ x_2&=\frac{1-4b+\sqrt{(20b^2-8b-3)}}{2}\\ x_3&=\frac{1+4b-\sqrt{(20b^2+8b-3)}}{2}\\ x_4&=\frac{1+4b+\sqrt{(20b^2+8b-3)}}{2} \end{array}$

To make these roots real valued we must have,

$20b^2-8b-3 \geq 0$ and $20b^2+8b-3 \geq 0$

these conditions and $b \geq 0$ can be simplified to,

$b\in [\frac{2+\sqrt{19}}{10},\infty)$

Now, as 2 of these roots are equal, there are 6 possible cases. Out of which $x_1=x_2$ results to,

$20b^2-8b-3 = 0$

which have only 1 root satisfying the condition,

$b=\frac{2+\sqrt{19}}{10}$

where, $a_1=(\frac{2+\sqrt{19}}{10})^2=\frac{23+\sqrt{304}}{100}$

Similarly $x_3=x_4$ results to,

$20b^2+8b-3 = 0$

which don't have any root satisfying the condition.

And other four cases can be combined as,

$\begin{array}{cc} (x_3-x_1)(x_3-x_2)(x_4-x_1)(x_4-x_2)&=0\\ (x_3^2-(1-4b)x_3+1-b^2)(x_4^2-(1-4b)x_4+1-b^2)&=0 \end{array}$

On simplifying it gives,

$64b^2(1-b^2)=0$

which have only one root satisfying the condition which is, $b=1$ . Let $a_2=1$

So, $a_1+a_2=\frac{123+\sqrt{304}}{100}\\ \Rightarrow k=304,m=100,n=123\\ \Rightarrow k+m+n=527$

If the quartic polynomial $f(x) \; = \; (x^2 - x + a + 1)^2 - 4a(5x^2 - x + 1)$ is to have exactly three distinct real zeros, it must have a repeated real root $u$ . Thus $f(u) = f'(u) = 0$ . Since $0 \; = \; f'(u) \; = \; 2\big[(u^2 - u + 1)(2u - 1) - a(18u - 1)\big]$ it is clear that $u \neq \tfrac{1}{18}$ and that $a \; = \; \frac{(u^2 - u + 1)(2u-1)}{18u-1}$ But then $f(u)$ is equal to $\left[u^2-u+1 + \frac{(u^2-u+1)(2u-1)}{18u-1}\right]^2 - 4\frac{(u^2-u+1)(2u-1)}{18u-1}(5u^2-u+1)$ which simplifies to give $f(u) \; = \; -\frac{16(u^2-u+1)u^2(20u^2-4u-15)}{(18u-1)^2}$ and hence we deduce that $u^2(20u^2 - 4u - 15) \; = \; 0$ so there are only three values of $u$ to consider.

1. If $u=0$ then $a=1$ and $f(x) \; = \; (x^2 - x + 2)^2 - 4(5x^2 - x + 1) \; = \; x^2(x - 5)(x + 3)$ and hence $a=1$ is one of the values we want.

2. If $u=\tfrac{1}{10}(1 + 2\sqrt{19})$ then $a = \tfrac{1}{100}(23 - 4\sqrt{19})$ and $f(x) \; = \; (x - u)^2 \left(\tfrac{1}{100}(77 + 4 \sqrt{19}) + \tfrac{1}{5} (-9 + 2 \sqrt{19}) x + x^2\right)$ and the second quadratic factor has negative discriminant $-\tfrac{8}{5} (\sqrt{19}-2)$ , and so two complex roots. We are not interested in this case.

3. Finally, if $u=\tfrac{1}{10}(1 - 2\sqrt{19})$ then $a = \tfrac{1}{100}(23 + 4\sqrt{19})$ and $f(x) \; = \; (x - u)^2 \left(\tfrac{1}{100}(77 - 4 \sqrt{19}) - \tfrac{1}{5} (9 + 2 \sqrt{19}) x + x^2\right)$ and this second quadratic factor has has positive discriminant $\tfrac{8}{5} (\sqrt{19}+2)$ , and so two distinct real roots. Neither of them are equal to $u$ . Thus $a=\tfrac{1}{100}(23 + 4\sqrt{19})$ is the other value we want.

Since $1 + \tfrac{1}{100}(23 + 4\sqrt{19}) \; = \; \tfrac{1}{100}(123 + 4\sqrt{19})\; = \; \tfrac{1}{100}(123 + \sqrt{304})$ the answer we want is $123 + 304 + 100 = 527$ .

Since $f(u)$ is to have exactly three real roots, the sensible way forward is to look for the repeated root, particularly since the condition $f'(u)=0$ is linear in $a$ , and hence can be solved to express $a$ in terms of $u$ . The condition $f(u)=0$ then becomes an equation in $u$ alone which (thankfully) simplifies considerably. Considering the three cases is just a matter of calculation.

Before I start presenting my solution, I would like to declare that this question is similar to the question Exactly 3 Real Solutions Allowed that was posted a few weeks back. I would also like to credit Ivan K. as I would be using a similar method as him (as he did in the other question).

Motivation When we see such questions about real solutions, we intuitively get reminded of the discriminant.Hence, we can manipulate the given equation to form $2$ separate quadratic equation. In our case, we want to find the value of $a$ such that the above equation has exactly $3$ distinct real solutions. This is only possible with $2$ cases:

1. Either one of the $2$ quadratic we have obtained (later) has a double root
2. (Or) The two quadratics share a common root

* Solution * So let's begin!

$(x^{2} -x +a -1)^{2} = 4a(5x^{2} --x +1)$ Let $n = x^{2} - x +1$ to simplify the manipulation.

Then, we observe that:

$(n+a)^{2} = 4a(n+4x^{2})$

$n^{2} + 2an + (a)^{2} = 4an + 16ax^{2}$

$n^{2} - 2(a)n +(a)^{2} = 16ax^{2}$

$(n-a)^{2} = (4x\sqrt{a})^{2}$

$x^{2} -x +1 -a = 4x\sqrt{a}$ or $-4x\sqrt{a}$

Let $\sqrt{a}$ be $u$ . Note that $u \geq 0$ .

$x^{2} -x+1 - u^{2} = 4xu$ or $-4xu$

Then, $x^{2} - (1+4u)x + (1-u^{2}) = 0$ --------------- (1)

or $x^{2} - (1-4u)+(1-u^{2})=0$ --------------- (2)

Now let's split our workings into 2 separate cases (as mentioned earlier):

• Case 1 : Either one of the 2 quadratics obtained above have a double root. This means that the discriminant of each quadratic must be $0$ .

Suppose the first equation (1) has double roots, then its discriminant

$(1+4u)^{2} - 4(1-u^{2}) = 0$

$1+ 8u + 16u -4 +4u^{2} = 0$

$20u^{2} +8u -3 = 0$

Using the quadratic formula, we get

$u = \frac{-2 \pm \sqrt{19}}{10}$ but note that since $u \geq 0$ , $u = \frac{\sqrt{19}-2}{10}$ .

Using the same method for supposing that the second equation has double roots, we obtain that $u = \frac{2+\sqrt{19}}{10}$ .

Now, we have to check each of our results (the 2 values of $u$ obtained earlier) by substituting them into equations (1) and (2):

For the first result, equation (1) becomes $x^{2} - (1+\frac{4\sqrt{19}-8}{10})x + (1-(\frac{\sqrt{19}-2}{10})^{2} = 0$ and equation (2) becomes $x^{2} - (1-\frac{4\sqrt{19}-8}{10})x + (1-(\frac{\sqrt{19}-2}{10})^{2} = 0$ .

Using the discriminant method for the above equations, you will notice that the first equation has $2$ real roots but the second equation does not have any real roots. Hence, $u \neq \frac{\sqrt{19}-2}{10}$ .

We continue with the same method with our second result, and we will find that the first equation has $2$ roots while the second equation has double roots, satisfying the conditions. Hence, $u =\frac{2+\sqrt{19}}{10}$ .

We are done with our first case, so let's proceed on with Case 2:

• Case 2 : The two quadratic equations share a common root.

Hence subtracting equations (1) and (2), we will obtain $-8ux = 0$ . From here, we know that either $u = 0$ or $x=0$ (or both).

Substituting $u=0$ into the given equation in the question, we will get $x^2 +4x +136 = 0$ which doesn't have real roots (we know this using the discriminant method).

Substituting $x=0$ into equation (1), we get $1-u^{2} = 0$ . Since $u\geq 0$ , $u=1$ .

Hence, possible answers for $u$ are $1$ and $\frac{2+\sqrt{19}}{10}$ . Since $a = u^{2}$ , the sum of $a$ is just $(1)^{2} + (\frac{2+\sqrt{19}}{10})^{2} = \frac{123+\sqrt304}{100}$ . Hence, $n = 123$ , $k = 304$ and $m = 100$ .

As such, we have arrived at the last step: $n+k+m = 123+304+100 = \boxed{527}$