Blog: `(x + 2)/(x(x^2 - 9))` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Tuesday, May 3, 2016

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{x}+{2}}}{{{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

Now let $\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}+{3}}}+\frac{{C}}{{{x}-{3}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{A}{\left({x}+{3}\right)}{\left({x}-{3}\right)}+{B}{\left({x}\right)}{\left({x}-{3}\right)}+{C}{\left({x}\right)}{\left({x}+{3}\right)}}}{{{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{A}{\left({{x}}^{{2}}-{9}\right)}+{B}{\left({{x}}^{{2}}-{3}{x}\right)}+{C}{\left({{x}}^{{2}}+{3}{x}\right)}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle\therefore{\left({x}+{2}\right)}={A}{\left({{x}}^{{2}}-{9}\right)}+{B}{\left({{x}}^{{2}}-{3}{x}\right)}+{C}{\left({{x}}^{{2}}+{3}{x}\right)}$

$\displaystyle{x}+{2}={A}{{x}}^{{2}}-{9}{A}+{B}{{x}}^{{2}}-{3}{B}{x}+{C}{{x}}^{{2}}+{3}{C}{x}$

$\displaystyle{x}+{2}={\left({A}+{B}+{C}\right)}{{x}}^{{2}}+{\left(-{3}{B}+{3}{C}\right)}{x}-{9}{A}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}+{C}={0}$

$\displaystyle-{3}{B}+{3}{C}={1}$

$\displaystyle-{9}{A}={2}$

Solve the above three equations to get the values of A,B and C,

$\displaystyle{A}=-\frac{{2}}{{9}}$

Bach substitute the value of A in the first equation,

$\displaystyle-\frac{{2}}{{9}}+{B}+{C}={0}$

$\displaystyle{B}+{C}=\frac{{2}}{{9}}$

From the above equation ,express C in terms of B

$\displaystyle{C}=\frac{{2}}{{9}}-{B}$

Substitute the expression of C in the second equation,

$\displaystyle-{3}{B}+{3}{\left(\frac{{2}}{{9}}-{B}\right)}={1}$

$\displaystyle-{3}{B}+\frac{{2}}{{3}}-{3}{B}={1}$

$\displaystyle-{6}{B}={1}-\frac{{2}}{{3}}$

$\displaystyle-{6}{B}=\frac{{1}}{{3}}$

$\displaystyle{B}=-\frac{{1}}{{18}}$

Now plug the value of...

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{{x}+{2}}}{{{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

Now let $\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}+{3}}}+\frac{{C}}{{{x}-{3}}}$

$\displaystyle\therefore{\left({x}+{2}\right)}={A}{\left({{x}}^{{2}}-{9}\right)}+{B}{\left({{x}}^{{2}}-{3}{x}\right)}+{C}{\left({{x}}^{{2}}+{3}{x}\right)}$

$\displaystyle{x}+{2}={A}{{x}}^{{2}}-{9}{A}+{B}{{x}}^{{2}}-{3}{B}{x}+{C}{{x}}^{{2}}+{3}{C}{x}$

$\displaystyle{x}+{2}={\left({A}+{B}+{C}\right)}{{x}}^{{2}}+{\left(-{3}{B}+{3}{C}\right)}{x}-{9}{A}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}+{C}={0}$

$\displaystyle-{3}{B}+{3}{C}={1}$

$\displaystyle-{9}{A}={2}$

Solve the above three equations to get the values of A,B and C,

$\displaystyle{A}=-\frac{{2}}{{9}}$

Bach substitute the value of A in the first equation,

$\displaystyle-\frac{{2}}{{9}}+{B}+{C}={0}$

$\displaystyle{B}+{C}=\frac{{2}}{{9}}$

From the above equation ,express C in terms of B

$\displaystyle{C}=\frac{{2}}{{9}}-{B}$

Substitute the expression of C in the second equation,

$\displaystyle-{3}{B}+{3}{\left(\frac{{2}}{{9}}-{B}\right)}={1}$

$\displaystyle-{3}{B}+\frac{{2}}{{3}}-{3}{B}={1}$

$\displaystyle-{6}{B}={1}-\frac{{2}}{{3}}$

$\displaystyle-{6}{B}=\frac{{1}}{{3}}$

$\displaystyle{B}=-\frac{{1}}{{18}}$

Now plug the value of B in the expression of C,

$\displaystyle{C}=\frac{{2}}{{9}}-{\left(-\frac{{1}}{{18}}\right)}$

$\displaystyle{C}=\frac{{2}}{{9}}+\frac{{1}}{{18}}$

$\displaystyle{C}=\frac{{{2}\cdot{2}+{1}}}{{18}}$

$\displaystyle{C}=\frac{{5}}{{18}}$

$\displaystyle\therefore\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}=-\frac{{2}}{{{9}{x}}}-\frac{{1}}{{{18}{\left({x}+{3}\right)}}}+\frac{{5}}{{{18}{\left({x}-{3}\right)}}}$

Now let's check the result,

RHS= $\displaystyle-\frac{{2}}{{{9}{x}}}-\frac{{1}}{{{18}{\left({x}+{3}\right)}}}+\frac{{5}}{{{18}{\left({x}-{3}\right)}}}$

$\displaystyle=\frac{{-{2}\cdot{2}{\left({x}+{3}\right)}{\left({x}-{3}\right)}-{x}{\left({x}-{3}\right)}+{5}{x}{\left({x}+{3}\right)}}}{{{18}{x}{\left({x}+{3}\right)}{\left({x}-{3}\right)}}}$

$\displaystyle=\frac{{-{4}{\left({{x}}^{{2}}-{9}\right)}-{\left({{x}}^{{2}}-{3}{x}\right)}+{5}{\left({{x}}^{{2}}+{3}{x}\right)}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{-{4}{{x}}^{{2}}+{36}-{{x}}^{{2}}+{3}{x}+{5}{{x}}^{{2}}+{15}{x}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{{18}{x}+{36}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{{18}{\left({x}+{2}\right)}}}{{{18}{x}{\left({{x}}^{{2}}-{9}\right)}}}$

$\displaystyle=\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$

=LHS

Hence it is verified.

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Tuesday, May 3, 2016

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Tuesday, May 3, 2016

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\frac{{{x}+{2}}}{{{x}{\left({{x}}^{{2}}-{9}\right)}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?