= Solution
At the <tricritical point>, both the quadratic and quartic even coefficients vanish, while the sextic coefficient remains positive. Two even control parameters must therefore be tuned, in addition to setting the symmetry-breaking <conjugate field> to zero. The <continuous phase transition> line and <first-order phase transition> line meet, the coexistence <order parameter> jump shrinks to zero, and the <tricritical wings> end there. The balance governing the equation of state changes from quartic to sextic stabilization.
Along the tricritical thermal path $u=0$, $r=a t$, $h=0$, minimizing $V$ gives $M^4=-r/v$ on the ordered side. At $r=u=0$, the equation of state is $h=vM^5$. Consequently
$$
\boxed{\beta_{\mathrm{tri}}=\frac14,\qquad \delta_{\mathrm{tri}}=5.}
$$
These <mean-field critical exponents> require the tuned tricritical path; a path with fixed $u>0$ instead has the ordinary exponents derived above. The corresponding equilibrium <Landau free energy> is $f_{\min}=-(-r)^{3/2}/(3\sqrt v)$ below the transition. Its first thermal derivative is continuous but its second derivative diverges, giving the tricritical <mean-field approximation> a heat-capacity exponent $\alpha=1/2$. Fluctuations are negligible asymptotically only above the tricritical <upper critical dimension> three; at three dimensions logarithmic corrections can accompany these powers.
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