Tangent To Quadratic And Cubic Polynomials

Let us say that the common tangent line is tangent to $x^2$ at $(a,a^2)$ and to $x^3$ at $(b,b^3)$ . Then by definition of the derivative,

Slope of line tangent to $x^2$ at $x=a$ $\implies 2a$ .

Slope of line tangent to $x^3$ at $x=b$ $\implies 3a^2$ .

Therefore the tangent line of $x^2$ is $\implies \frac{y-a^2}{x-a}=2a \implies \boxed{y - 2ax + a^2 = 0}$ , and that of $x^3$ is $\implies \frac{y-b^3}{x-b}=3b^2 \implies \boxed{y - 3b^2x + 2b^3=0}$ .

Clearly, these lines must be the same if they are tangent to both the curves. Hence, comparing the lines:

$\implies 2a=3b^2, \implies a^2=2b^3$ .

Solving the set of equations, we have two values of $b=0, \frac{8}{9}$ . Substituting $b=0$ in the tangent line of $x^3$ gives $y=0$ (As I had stated before) and $b=\frac{8}{9}$ gives $\boxed{729y-1728x+1024=0}$ , which is the required tangent line equation.

Hence, $|a|+|b|+|c|=|729|+|-1728|+|1024|=\boxed{3481}$

Here's a non-calculus approach: We can find the equation of the desired straight line by solving for their respective polynomial discriminants.

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To avoid the abuse of notations, I will define the equation of the straight line that is tangent to both the curves as $y = Ax + B$ , where $(A,B) \ne (0,0)$ .

Hence, both the equations $\begin{cases} x^2 = Ax + B \\ x^3 = Ax + B \end{cases}$ has exactly one real root because there is only one intersection point.

Because there's only one real solution, the quadratic discriminant of the quadratic equation $x^2 = Ax + B$ must be equal to 0:

$(-A)^2 -4(1)(-B) = 0 \qquad \Leftrightarrow \qquad B = \dfrac{A^2}4$

Similarly, the cubic discriminant of the cubic equation $x^3 = Ax + B \Rightarrow x^3 - Ax - \dfrac{A^2}4 = 0$ must be 0:

$\begin{aligned} && (0^2)(-A) - 4(1)(-A)^3 - 4(0)^3\left( -\dfrac{A^2}4 \right) - 27(1)^2 \left( -\dfrac{A^2}4\right)^2 + 18(1)(0)(-A)\left( -\dfrac{A^2}4\right) = 0 \\ &\Rightarrow & 4A^3 + \dfrac{27}{16}A^4 = 0 \\ &\Rightarrow & A = -\dfrac{64}{27}, 0 \\ &\Rightarrow & (A,B) = \left( -\dfrac{64}{27},\dfrac{1024}{729} \right) , (0,0) \\ \end{aligned}$

But since we're looking for a straight line other than $y=0$ , then $(A,B) \ne (0,0 )$ as stated above. Hence $(A,B) = \left( -\dfrac{64}{27},\dfrac{1024}{729} \right)$ only. And so the desired equation is

$y = -\dfrac{64}{27} x + \dfrac{1024}{729} \qquad \Leftrightarrow \qquad 729y - 1728x + 1024 = 0$

with the answer $|729| + |-1728 | + |1024| = \boxed{3481 }$ .

Let ${\bf f(x) = x^2 }$ and ${\bf g(x) = x^3 }$

${\bf \frac{d}{dx}(f(x))|_{x = a} = 2a }$ and ${\bf \frac{d}{dx}(g(x))|_{x = b} = 3b^2 }$ $$

Since both functions have a common tangent ${\bf \implies 2a = 3b^2 \implies a = \frac{3}{2}b^2 }$

Let ${\bf A:(\frac{3}{2} b^2, \frac{9}{4} b^4) }$ ${\bf B:(b,b^3) }$

slope ${\bf m = 3b^2 = \frac{1}{2} b^2(\frac{9b - 4}{3b - 2}) }$

${\bf b \ne 0 \implies 9b - 4 = 18b -12 \implies b = \frac{8}{9} \implies a = \frac{8}{27} \implies m = \frac{64}{27} }$ $$

Using ${\bf B:(\frac{8}{9}, \frac{512}{729}) \implies 1728x - 729y -1024 = 0 }$ $$ where, ${\bf a = 1728, b = -729, c = - 1024 }$ and ${\bf (|a|,|b|,|c|) = 1 \implies |a| + |b| + |c| = 3481}$