How to solve $6x^{2/3}-12-2x=0$?

Question

How to solve $6x^{2/3}-12-2x=0$? I need it to prove that the graphic of $y=6x^{2/3}$ never meets the graphic of function $y=12+2x$.

Answer 1

If $t=x^{1/3}$, the equation is $$ t^3-3t^2+6=0 $$ The only real root of this equation is negative. Indeed the derivative of $f(t)=t^3-3t^2+6$ is $f'(t)=3(t^2-2)$, that vanishes at $t=-\sqrt{2}$ and $t=\sqrt{2}$. Since $$ f(-\sqrt{2})=-2\sqrt{2}<0 \qquad f(\sqrt{2})=2\sqrt{2}>0 $$ the polynomial has a single real root, which is negative because $f(0)=6>0$.

Thus the curve $y=x^{2/3}$ and the line $y=12+2x$ intersect at a point with $x<0$.

If you consider $x^{2/3}$ only defined for $x\ge0$, then you're right, there is no intersection.

Answer 2

We have $x+6 = 3x^{2/3}$ i.e $(x+6)^3 = 27x^2$. This has a real solution. You won't be able to prove what you wish.

Answer 3

We have $6^3x^{2}=(12+2x)^3$. Use the Cardamo formula for cubic equation.

Answer 4

Let $x^{\frac{1}{3}}=y$.

Thus, $x>0$, $y>0$ and we need to solve $$y^3-3y^2+6=0.$$ But, by AM-GM $$y^3+6=2\left(\frac{y^3}{2}\right)+6\geq3\sqrt[3]{6\left(\frac{y^3}{2}\right)^2}>3y^2,$$ which says that our equation has no real roots.