How to solve the equation $3a^4=8a^3-16$?, where a is a positive number
I have tried using the quadric formula, but got to nothing useful.
I know the solution is $2$, but I don't know how to solve it. This is a part of a problem and I need it to solve the whole problem.
Hope one of you can help me! Thank you!
If you know that $a=2$ is the root then just divide $f(a)=3a^4-8a^3+16$ by $a-2$ and lower the degree. Actually $f(a)=3a^4-8a^3+16=(a-2)^2(3a^2+4a+4)$. The discriminant of $3a^2+4a+4$ is negative, so no roots except $a=2$.
If you don’t want to factor $f(a)=3a^4-8a^3+16$, you can take its derivative: $$f’(a)=12a^3-24a^2=12a^2(a-2).$$
This means that $f(a)$ has a global minimum at $a=2$. Since $f(2)=0$, it’s the only solution.
$$3a^4-8a^3+16=0$$ $$3a^4-6a^3-2a^3+16=0$$ $$3a^3(a-2)-2(a^3-8)=0$$ $$(a-2)(3a^3-2(a^2+2a+4))=0$$ $$(a-2)(3a^3-2a^2-4a-8)=0$$ $$(a-2)(a^3-8+2a^3-2a^2-4a)=0$$ $$(a-2)(a^3-8+2a(a^2-a-2))=0$$ $$(a-2)^2(a^2+2a+4+2a(a+1))=0$$ $$(a-2)^2(3a^2+4a+4)=0$$ Hence, $a_{1,2}=2$ and $a_{3,4}=\frac{-2\pm\sqrt2 i}{3}$ by quadratic formula.
