Find all quadratic polynomials $f(x)$,$g(x)$ and $h(x)$ such that the polynomial $ f(g(h(x)))=0 $ has roots $1;2;..;8$
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$ f(g(h(x)))=0 $ has roots $1;2;..;8$
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algebra-precalculus
polynomials
roots
quadratics
1 Answers
3
We have $f(z)=0$ for at most two values $z_1,z_2$ of $z$. And we have $g(y)=z_i$ for at most two values $y_{i,1}, y_{i,2}$ of $y$. And we have $h(x)=y_{i,j}$ for at most two values $x_{i,j,1}, x_{i,j,2}$ of $x$. In order to have eight roots in total, "at most two" must in fact be "exactly two" in all levels, and the eight numbers $x_{1,1,1},x_{1,1,2},x_{1,2,1},\ldots, x_{2,2,2}$ must be $1,2,3,4,5,6,7,8$ in some order. Note that $x_{i,j,1}$ and $x_{i,j,2}$ are always symmetric about the vertex of the parabola described by $h$. Therefore, this vertex must be at $4\frac12$ and so $h(x)=a(2x-9)^2+b$. Then we have $$h(1)=h(8)=49a+b, h(2)=h(7)=25a+b,h(3)=h(6)=9a+b, h(4)=h(5)=a+b.$$ Now we need that the four roots $y_{1,1},y_{1,2},y_{2,1},y_{2,2}$ of $f(g(y))$ must be $a+b,9a+b,25a+b,49a+b$ in some order. As above, $y_{i,1}$ and $y_{i,2}$ are symmetric about the vertex of $g$. But that would require $\frac{(49a+b)+(a+b)}2=\frac{(9a+b)+(25a+b)}2$, i.e., $a=0$ and $h$ is not quadrtatic after all. We conclude that no quadratic polynomials $f,g,h$ as desired exist.