Question

If sum of values of ${}^{\prime }{x}^{\prime }$ of equation $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$ is $\alpha$ . Then find $|\alpha |$.

Answer 1

$x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$

In this case we can see that,

$\frac{2}{3}=2\left(\frac{1}{3}\right)$

and so one of the exponents is twice the other so it looks like we’ve got an equation that is reducible to quadratic in form. The substitution will then be

$u=x^{\frac{1}{3}}$

$\therefore$ $u^2=x^{\frac{2}{3}}$

Substituting this into the equation gives,

$u^2+u-2=0$

By using formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ we get factors,

$u=-1+\sqrt{3}$ and $u=-1-\sqrt{3}$

So, $x^{\frac{1}{3}}=-1+\sqrt{3}$

$\therefore$ $x=(-1+\sqrt{3}{)}^3=(-1{)}^3+(\sqrt{3}{)}^3+3(-1\left)\right(\sqrt{3}\left)\right(-1+\sqrt{3})$

$\therefore$ $x=-1+3\sqrt{3}-3\sqrt{3}(-1+\sqrt{3})=-10+6\sqrt{3}$ ----- ( 1 )

$\Rightarrow$ Now, $x^{\frac{1}{3}}=(-1-\sqrt{3})$

$x=(-1-\sqrt{3}{)}^3=(-1{)}^3-(\sqrt{3}{)}^3-3(-1\left)\right(\sqrt{3}\left)\right(-1-\sqrt{3})$

$\therefore$ $x=-10-6\sqrt{3}$ ------ ( 2 )

$\Rightarrow$ Sum of values of $x$ $=(-10+6\sqrt{3})+(-10-6\sqrt{3})$ [ From ( 1 ) and ( 2 ) ]

$\therefore$ $\alpha =-20$

$\therefore$ $|\alpha |=|-20|=20$