Say you have the equation $y = 9x^4 + 7x^2 + 2$. There are multiple ways of finding the roots of this equation, but one of them is to let $u = x^2$, then re-write the equation as $y = 9u^2 + 7u + 2$, solve that quadratic, and then set the $u$ values equal to $x^2$ and solve. But if I am only trying to determine if it has any real roots, then could I just say that because the quadratic in terms of $u$ has no real roots, then the original equation has no real roots aswell? Or is this wrong to assume?
Because if you have the equation $y = 9x^4 + 7x^2$, and follow the method above, then the original equation has no real roots, but the quadratic equation ($9u^2 + 7u$) does have real roots. Is this becuase there is no constant value in the equation unlike in the first? Is there a case where the quartic does have real roots but the hidden quadratic doesn't, even with a constant?
My question is, if you have a quartic equation in the form $ax^4 + bx^2 + c $ where $a, b, c ≠ 0$, and you let $u = x^2$, then do both equations in terms of $x$ and in terms of $u$ follow the same characteristic in terms of roots (both have no real roots, both have real roots), or can you assume that if the "hidden quadratic"/new equation has no real roots, then the quartic equation also doesn't have any real roots.