Quartic equation $ x^4-mx^2-4=0$ has root $\sqrt[4]{3-2\sqrt2}+\sqrt[4] {3+2\sqrt2}$. Find $m$

Question

Find $m$ knowing that $$x^4-mx^2-4=0$$ has $$ \sqrt[4]{3-2\sqrt2}+\sqrt[4] {3+2\sqrt2}$$ as root.

I was thinking that if it has this root it must have $ \sqrt[4]{3-2\sqrt2}-\sqrt[4] {3+2\sqrt2}$ too. Am I wrong? I tried solving by replacing $x^2$ with t too but it doesn't lead anywhere cause the roots are kind of weird looking.

Answer 1

Note that you can simplify root as $$\sqrt [ 4 ]{ (3-2\sqrt { 2 } ) } +\sqrt [ 4 ]{ (3+2\sqrt { 2 } ) } =\sqrt [ 4 ]{ { \left( \sqrt { 2 } -1 \right) }^{ 2 } } +\sqrt [ 4 ]{ { \left( \sqrt { 2 } +1 \right) }^{ 2 } } =\\ =\sqrt { \sqrt { 2 } -1 } +\sqrt { \sqrt { 2 } +1 } $$

Answer 2

Id est, $(2+2\sqrt2)^2-m(2+2\sqrt2)-4=0$, which gives $m=4$.

$ \sqrt[4]{3-2\sqrt2}-\sqrt[4] {3+2\sqrt2}$ is a root, of course.

Answer 3

If $\sqrt[4]{3-2\sqrt{2}}+\sqrt[4]{3+2\sqrt{2}}$ is a root of $x^4-mx^2-4=0$ then $(\sqrt[4]{3-2\sqrt{2}}+\sqrt[4]{3+2\sqrt{2}})^2$ is a root of $t^2-mt-4=0$ so we get that $$t_1=(\sqrt[4]{3-2\sqrt{2}}+\sqrt[4]{3+2\sqrt{2}})^2=\sqrt{3-2\sqrt{2}}+2\sqrt[4]{9-8}+\sqrt{3+2\sqrt{2}}\\=\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}+2$$ If we denote $\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=p$ then $$p^2=3-2\sqrt{2}+2+3+2\sqrt{2}=8$$ so $p=2\sqrt{2}$ from which we get that $$t_1=2\sqrt{2}+2$$ plugging it back in $t^2-mt-4=0$ we get $$(2+2\sqrt{2})^2-m(2\sqrt{2}+2)-4=0\\4(\sqrt{2}+1)^2-2m(\sqrt{2}+1)-4=0\\2(3+2\sqrt{2})-m(\sqrt{2}+1)-2=0\\4+4\sqrt{2}=m(\sqrt{2}+1)\\m=4$$

Answer 4

Rewrite $x^4-mx^2-4=0$ as $$(x^2-\frac m2)^2=4+\frac {m^2}4$$ which leads to the solution $$x =\sqrt{\sqrt{4+\frac{m^2}4}+\frac m2 }$$

Then, compare with the given root $$ \sqrt[4]{3-2\sqrt2}+\sqrt[4] {3+2\sqrt2} = \sqrt{\sqrt2-1}+\sqrt {\sqrt2+1}= \sqrt{\sqrt8+2} $$ to arrive at $m=4$.