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If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.

We may now assume $m\neq 0$. After dividing your sextic by $m^6$, and taking $y:=x/m$ and $K:=L/m^6$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).

The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run

Manipulate[ Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]

in Mathematica, you can visually see how the two roots behave, as $K$ varies.

Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of $$ w^3+5w^2+3w+J $$ when $J\leq 0$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]

you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $J\leq 0$.

Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).

If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.

We may now assume $m\neq 0$. After dividing your sextic by $m^6$, and taking $y:=x/m$ and $K:=L/m^6$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).

The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run

Manipulate[Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]

in Mathematica, you can visually see how the two roots behave, as $K$ varies.

Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of $$ w^3+5w^2+3w+J $$ when $J\leq 0$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]

you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $J\leq 0$.

Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).

If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.

We may now assume $m\neq 0$. Then take $K:=L/m^6$. After dividing your sextic by $m^6$, and taking $y:=x/m$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).

The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run

Manipulate[ Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]

in Mathematica, you can visually see how the two roots behave, as $K$ varies.

Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of $$ w^3+5w^2+3w+J $$ when $J\leq 0$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]

you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $J\leq 0$.

Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).

If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.

We may now assume $m\neq 0$. After dividing your sextic by $m^6$, and taking $y:=x/m$ and $K:=L/m^6$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).

The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run

Manipulate[ Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]

in Mathematica, you can visually see how the two roots behave, as $K$ varies.

Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of $$ w^3+5w^2+3w+J $$ when $J\leq 0$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]

you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $J\leq 0$.

Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).

If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.

We may now assume $m\neq 0$. Then take $K:=L/m^6$. After dividing your sextic by $m^6$, and taking $y:=x/m$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).

The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run

Manipulate[ Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]

in Mathematica, you can visually see how the two roots behave, as $K$ varies.

Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of $$ w^3+5w^2+3w+J $$ when $J\leq 0$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]

you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your positive real root, because that expression is continuous as a function in $J\leq 0$.

Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).

If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.

We may now assume $m\neq 0$. Then take $K:=L/m^6$. After dividing your sextic by $m^6$, and taking $y:=x/m$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).

The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run

Manipulate[ Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]

in Mathematica, you can visually see how the two roots behave, as $K$ varies.

Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of $$ w^3+5w^2+3w+J $$ when $J\leq 0$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]

you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $J\leq 0$.

Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).