smc

The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+a{x}^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+a{x}^2-bx-c$. Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$. Suppose we let $p$, $q$, and $r$ be the roots of this function, and let $x^3-u{x}^2+vx-w$ be the cubic polynomial with roots $p$, $q$, and $r$. $$\begin{array}{rl}(x-p)(x-q)(x-r)& =x^3-u{x}^2+vx-w\\ (x-p{)}^2(x-q{)}^2(x-r{)}^2& =x^6-10x^5+29x^4-4x^3+a{x}^2-bx-c=0\\ \sqrt{x^6-10x^5+29x^4-4x^3+a{x}^2-bx-c}& =x^3-u{x}^2+vx-w=0\end{array}$$ In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+a{x}^2-bx-c}$ we must first expand out the terms of $(x^3-u{x}^2+vx-w{)}^2$. $$(x^3-u{x}^2+vx-w{)}^2$$ $$=x^6-2u{x}^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2$$ [Quick note: Since we don't know $a$, $b$, and $c$, we really don't even need the last 3 terms of the expansion.] $$\begin{array}{rl}& 2u=10\\ u^2+2v& =29\\ 2uv+2w& =4\\ u& =5\\ v& =2\\ w& =-8\\ & \sqrt{x^6-10x^5+29x^4-4x^3+a{x}^2-bx-c}=x^3-5x^2+2x+8\end{array}$$ All that's left is to find the largest root of $x^3-5x^2+2x+8$. $$\begin{array}{rl}& x^3-5x^2+2x+8=(x-4)(x-2)(x+1)\\ & \boxed{4}\end{array}$$