$f=2x^{4}+4x^{3}+3x^{2}+bx+c$ has all real roots, find b,c (b,c are from R). Thanks a lot, I tried with substitution, I don't know, is there something with derivative? please help thanks
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2Well, between any two zeros of the function, there's a zero of the derivative, though I'm not sure this helps. – saulspatz Apr 20 '20 at 22:16
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2Is the exercise to find every pair $b,c$ for which $f$ has only real roots, or is it to find just one such pair? Also, please give us some context. Where did you come across this problem? What are your thoughts on the problem? What have you tried? Where are you getting stuck? – Ben Grossmann Apr 20 '20 at 22:18
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@saulspatz In this case it certainly helps, since that observation implies a condition on the discriminant of $f'$ and hence a condition on $b$ alone. – Travis Willse Apr 20 '20 at 22:29
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Let $g(x)=2x^4+4x^3+3x^2$. Then $g''(z)=24x^2+24x+6=6(2x+1)^2\geq0$. So $g$ is convex, and will intersect a line at most twice. In order for $f(x)$ to have only real roots, the line $y=-bx-c$ will have to be tangent to the graph of $g$. But the tangent will not intersect the graph in a second point, so if $f$ has $4$ real zeros, it must have a quadruple zero, that is $f$ and its first three derivatives must vanish at some point.
The computation above shows that this point must be $x=-\frac12$ and we must have $$f(x)=2\left(x+\frac12\right)^4.$$
I haven't carried it past this point. I leave it to you.
saulspatz
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$f$ has $4$ (not necessarily distinct) real roots, say $r, s, t, u$. Hence, $f$ can be factorized this way : $f = 2 (x-r) (x -s) (x-t) (x - u)$. If you developp this expression, and compare term by term, you'll get that the roots satisfy he following system of equations (edited 4 equations, as pointed out by @saulspatz) : $$\left\{ \begin{array}{ccl} c & = & 2 rstu \\ b & = & -2(rst + rsu + rtu + stu) \\ 3 & = & 2 (rs + rt + ru + st + su + tu) \\ 4 & = & -2(r + s + t + u) \end{array}\right. $$
Granted, this looks ugly, but only the last two lines represents a constraint. e.g. take $r = s = t = u = \frac{-1}{2}$, this satisfies the two last lines and hence you get a solution where $b = 1$ and $c = \frac{1}{8}$.
Olivier Roche
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1This should be a system of 4 equations: The third equation should have $3$ on the left-hand side (since it corresponds to the quadratic term), and the cubic term gives us the last of Vieta's formulas: $4 = -2(r + s + t + u)$ – Travis Willse Apr 20 '20 at 22:53
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yes, something is not right, I don't know if you can make it with viete, sorry for that – Not important Apr 21 '20 at 01:26
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2Possibly. That's what Olivier's answer does, except for the transcription error I mentioned above. saulsplatz' answer is possibly the best elementary answer. Another option is to use saulsplatz' comment under the original question, which shows that $f'$ must have three real roots. In terms of the variable $x = y - \frac{1}{2}$, the derivative is $f' = 8y^3 + (b - 1)$. But (for any $b$) this function is nondecreasing, so it has three real roots iff all of those roots coincide, i.e., if $b - 1 = 0$. Substituting in the original equation gives $f = 2 y^4 + \left(c - \frac{1}{8}\right)$. – Travis Willse Apr 21 '20 at 01:50
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$f(x) = 2x^4+4x^3+3x^2+bx+c$
If $f(x)$ has all $4$ roots to be real and complex, then it's discriminate $$\Delta = (-72)b^2+176b^3-108b^4-864c-2304bc+1536b^2c+2304c^2-3072bc^2+2048c^3$$ must be real, $\Delta > 0$ Also if you divide $f(x)$ by 2, and make a translation $x = y-\frac{1}{2}$
$2x^4+4x^3+3x^2+bx+c = 0$
$x^4+2x^3+\frac{3/2}x^2+\frac{b/2}x+\frac{c/2} = 0$
Say $x = y-\frac{1}{2}$
The equation is depressed that suddenly appear to be
$y^4+(\frac{b}{2}-\frac{1}{2})y+\frac{c}{2}-\frac{b}{4}+\frac{3}{16} = 0$
Notice that the $y^2$ automatically vanishes
Defined $D = 512c-256b+192$
For the polynomial to therefore have real distinct root $D < 0$ and $\Delta > 0$
