Notes: How do I solve for x when `y = x^(1/3)` and `x^(1/3) - 2x^(-1/3) = 1` And how do I simplify: `(sqrt(3))^-3 + (sqrt(3))^-2 + (sqrt(3))^-1 +...

Monday, 14 October 2013

How do I solve for x when $\displaystyle{y}={{x}}^{{\frac{{1}}{{3}}}}$ and $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}-{2}{{x}}^{{-\frac{{1}}{{3}}}}={1}$ And how do I simplify: $\displaystyle{{\left(\sqrt{{{3}}}\right)}}^{{-{{3}}}}+{{\left(\sqrt{{{3}}}\right)}}^{{-{{2}}}}+{{\left(\sqrt{{{3}}}\right)}}^{{-{{1}}}}+\ldots$

We are asked to solve $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}-{2}{{x}}^{{-\frac{{1}}{{3}}}}={1}$ with the hint to let $\displaystyle{y}={{x}}^{{\frac{{1}}{{3}}}}$ :

If we let $\displaystyle{y}={{x}}^{{\frac{{1}}{{3}}}}$ we get:

$\displaystyle{y}-{2}{{y}}^{{-{1}}}={1}$ since $\displaystyle{{x}}^{{-\frac{{1}}{{3}}}}={{\left({{x}}^{{\frac{{1}}{{3}}}}\right)}}^{{-{1}}}$

or $\displaystyle{y}-\frac{{2}}{{y}}={1}$ Multiplying by y yields:

$\displaystyle{{y}}^{{2}}-{2}={y}=\Rightarrow{{y}}^{{2}}-{y}-{2}={0}$

This factors as $\displaystyle{\left({y}-{2}\right)}{\left({y}+{1}\right)}={0}$ so we get $\displaystyle{y}={2}$ or $\displaystyle{y}=-{1}$ :

If y=2 then $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}={2}=\Rightarrow{x}={8}$

If y=-1 we get $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}=-{1}=\Rightarrow{x}=-{1}$

Checking the solutions we see that $\displaystyle{{8}}^{{\frac{{1}}{{3}}}}-{2}{{\left({8}\right)}}^{{-\frac{{1}}{{3}}}}={2}-\frac{{2}}{{2}}={1}$ and

$\displaystyle{{\left(-{1}\right)}}^{{\frac{{1}}{{3}}}}-\frac{{2}}{{{\left(-{1}\right)}}^{{\frac{{1}}{{3}}}}}=-{1}-\frac{{2}}{{-{1}}}=-{1}+{2}={1}$ ...

We are asked to solve $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}-{2}{{x}}^{{-\frac{{1}}{{3}}}}={1}$ with the hint to let $\displaystyle{y}={{x}}^{{\frac{{1}}{{3}}}}$ :

If we let $\displaystyle{y}={{x}}^{{\frac{{1}}{{3}}}}$ we get:

$\displaystyle{y}-{2}{{y}}^{{-{1}}}={1}$ since $\displaystyle{{x}}^{{-\frac{{1}}{{3}}}}={{\left({{x}}^{{\frac{{1}}{{3}}}}\right)}}^{{-{1}}}$

or $\displaystyle{y}-\frac{{2}}{{y}}={1}$ Multiplying by y yields:

$\displaystyle{{y}}^{{2}}-{2}={y}=\Rightarrow{{y}}^{{2}}-{y}-{2}={0}$

This factors as $\displaystyle{\left({y}-{2}\right)}{\left({y}+{1}\right)}={0}$ so we get $\displaystyle{y}={2}$ or $\displaystyle{y}=-{1}$ :

If y=2 then $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}={2}=\Rightarrow{x}={8}$

If y=-1 we get $\displaystyle{{x}}^{{\frac{{1}}{{3}}}}=-{1}=\Rightarrow{x}=-{1}$

Checking the solutions we see that $\displaystyle{{8}}^{{\frac{{1}}{{3}}}}-{2}{{\left({8}\right)}}^{{-\frac{{1}}{{3}}}}={2}-\frac{{2}}{{2}}={1}$ and

$\displaystyle{{\left(-{1}\right)}}^{{\frac{{1}}{{3}}}}-\frac{{2}}{{{\left(-{1}\right)}}^{{\frac{{1}}{{3}}}}}=-{1}-\frac{{2}}{{-{1}}}=-{1}+{2}={1}$ as required.

The solutions are x=8 and x=-1. The graph of $\displaystyle{y}={{x}}^{{\frac{{1}}{{3}}}}-{{x}}^{{-\frac{{1}}{{3}}}};{y}={1}$ :

-----------------------------------------------------------------------------

One way to simplify $\displaystyle{\sqrt{{{3}}}}^{{-{3}}}+{\sqrt{{{3}}}}^{{-{2}}}+{\sqrt{{{3}}}}^{{-{1}}}+{\sqrt{{{3}}}}^{{0}}+{\sqrt{{{3}}}}^{{1}}+{\sqrt{{{3}}}}^{{2}}+{\sqrt{{{3}}}}^{{3}}$ is to use the substitution $\displaystyle{y}=\sqrt{{{3}}}$ to get the expression:

$\displaystyle{{y}}^{{-{3}}}+{{y}}^{{-{2}}}+{{y}}^{{-{1}}}+{{y}}^{{0}}+{{y}}^{{1}}+{{y}}^{{2}}+{{y}}^{{3}}$

Factoring out a common $\displaystyle{{y}}^{{-{3}}}$ we get:

$\displaystyle{{y}}^{{-{3}}}{\left[{1}+{y}+{{y}}^{{2}}+{{y}}^{{3}}+{{y}}^{{4}}+{{y}}^{{5}}+{{y}}^{{6}}\right]}$

The expression in brackets can be rewritten recognizing the identity:

$\displaystyle{{y}}^{{7}}-{1}={\left({y}-{1}\right)}{\left({1}+{y}+{{y}}^{{2}}+{{y}}^{{3}}+{{y}}^{{4}}+{{y}}^{{5}}+{{y}}^{{6}}\right)}$ , so the expression in the brackets becomes $\displaystyle\frac{{{{y}}^{{7}}-{1}}}{{{y}-{1}}}$

Substituting for y we get $\displaystyle{\sqrt{{{3}}}}^{{-{3}}}{\left[\frac{{{\sqrt{{{3}}}}^{{7}}-{1}}}{{\sqrt{{{3}}}-{1}}}\right]}$

Using the rules for simplifying radical expressions we end up with :

$\displaystyle\frac{{{40}\sqrt{{{3}}}}}{{9}}+\frac{{13}}{{3}}$

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)