Double Tangent

$y = ax+b$ $y = x^4 - 2x^3 -9x^2 + 2x +8$ Equating both, we get $ax+b = x^4 - 2x^3 -9x^2 + 2x +8$ $x^4 - 2x^3 -9x^2 + (2-a)x +8-b = 0$ Since the line is tangetial to the quartic curve at exactly 2 distinct points, the equation above can be express in the form of $(x-\alpha)^2(x-\beta)^2=0$ $[(x-\alpha)(x-\beta)]^2=0$ From this we know that LHS can be expressed in the form of a square of a degree 2 polynomial. So let LHS be $(px^2 + qx +r)^2$ $(px^2 + qx)^2 + r^2 + 2r(px^2 + qx)$ $(p^2x^4 + q^2x^2 + 2pqx^3 + r^2 + 2prx^2 + 2qrx$ $p^2x^4 + 2pqx^3 + (2pr + q^2)x^2 + 2qrx + r^2$ $\Rightarrow p^2x^4 + 2pqx^3 + (2pr + q^2)x^2 + 2qrx + r^2 = x^4 - 2x^3 -9x^2 + (2-a)x +8-b$ Comparing coefficient of $x^4$ , $p^2 = 1$ $p =\pm 1$ Comparing coefficient of $x^3$ , $2pq = -2$ $\pm2q = -2$ $q = \mp 1$ Comparing coefficient of $x^2$ , $2pr + q^2 = -9$ $\pm 2r + (\mp1)^2 = -9$ $r = \mp 5$ Comparing coefficient of $x$ , $2qr = 2 -a$ $2(\mp 1)(\mp 5) = 2 -a$ $10 = 2 -a$ $a = -8$ Comparing constants, $r^2 = 8 -b$ $25 = 8 - b$ $b = -17$ So $ab = (-8)(-17) = 136$

y=ax+b is a tangent to the curve y=x^4 + lower order terms at two points and this means that when we subtract the two functions of x we have two repeated roots as the tangent and the quartic equation touch twice. That means the difference between the two functions = ((x-c)^2)((x-d)^2) for some c and d. Now this is in fact an identity which means we are allowed to expand the RHS and equate coefficients of each power of x. So let us proceed, RHS= x^4 -2(c+d)x^3 +(c^2 + 4cd + d^2)x^2 -2cd(c+d)x + (cd)^2 = difference between quartic and tangent equation = x^4 - 2x^3 -9x^2 + (2-a)x + 8 - b. Now considering the coefficients of the x^3 terms, we have: c+d=1, which allows us to simplify the other terms in our expression in c,d and x. In particular note that considering the x^2 terms, c^2 + 4cd + d^2 = (c+d)^2 + 2cd = 1 + 2cd = -9, which yields cd=-5. This means we can simplify even further now the x terms and constants. Therefore 2-a = -2cd(c+d)=10 (from previous working) hence a=-8. And also (cd)^2 = 25 = 8 - b which implies b = -17. Therefore we have a = -8 and b = -17. This gives ab = 136. That is my opinion.

We are given the equation of the tangent as ${y}={ax+b}$ and that of the graph as ${y}={x^4}-{2x^3}-{9x^2}+{2x}+{8}$ subtracting the two equations we get ${x^4}-{2x^3}-{9x^2}+({2-a})x+({8-b}).....(1)$ We can see that the above expression as two repeated roots or it can also be written as $({x^2+px+q})^2$ expanding that we get ${x^4}+{2px^3}+({p^2+2q})x^2+{2pqx}+{q^2}.....(2)$ but .....(1) and .....(2) are equivalent. so comparing the coefficients we get ${2px^3}={-2x^3}$ from which ${p}={-1}$ $({p^2+2q})={-9}$ from which ${q}={-5}$ ${2pqx}=({2-a})x$ from which ${a}={-8}$ ${q^2}=({8-b})$ from which ${b}={-17}$ Hence ${ab}={-8\times{-17}}=\boxed{136}$

$$ Let $y_1 = ax+b$ and $\,y_2=x^4 - 2x^3 - 9x^2 + 2x + 8$ . Since $\,y_1$ is a tangent line of $\,y_2$ at two distinct points, then $\,y_2-y_1$ will have two roots. Strictly speaking $$ $\begin{aligned} (x^2+mx+n)^2 &= y_2-y_1\\ x^4+2mx^3 + (m^2+2n)x^2+2mnx+n^2&=x^4 - 2x^3 - 9x^2 + (2-a)x + (8-b) \end{aligned}$ $$ By comparing coefficients left-side and right-side equations, we'll obtain $\,a=-8\,$ and $\,b=-17\,$ . Hence, $\,a\cdot b=\boxed{136}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

I will do a linear transformation that takes $y = ax+b$ onto the $x$ -axis, so that we end up with a quartic that touches the x-axis in two distinct points.

If we make the change of variables $\begin{aligned} X &= x \\ Y &= ax + b - y \end{aligned}$ then the line becomes $Y=0$

and the quartic is $\begin{aligned} Y &= ax + b - (x^4 - 2x^3 - 9x^2 + 2x + 8)\\ &= X^4 - 2X^3 - 9X^2 + (2-a)X + (8-b) \end{aligned}$

Now, I am going to use the result "If a quartic is tangent to the $x$ -axis in two places then it has two roots, each of multiplicity $2$ ". I don't know why this is true, but it has come up before on Brilliant on a similar problem, and the solvers assumed it without explaining it.

Given this, we can write $Y = (X - \alpha)^2 (X - \beta)^2$

Expanding this out gives:

$Y = X^4 - 2(\alpha + \beta)X^3 + ({\alpha}^2 + {\beta}^2 + 4\alpha \beta)X^2 - ( 2\alpha \beta (\alpha + \beta))X + (\alpha \beta)^2$

Equating coefficients with the equation of the quartic earlier gives the following relations: $\begin{aligned} \alpha + \beta &= 1 \\ {\alpha}^2 + {\beta}^2 + 4\alpha \beta &= -9 \\ 2 \alpha \beta &= 2-a \\ (\alpha \beta)^2 &= 8-b \end{aligned}$