Blog: `1/(4x^2 - 9)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Monday, November 2, 2015

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}=\frac{{1}}{{{\left({2}{x}+{3}\right)}{\left({2}{x}-{3}\right)}}}$

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}=\frac{{A}}{{{2}{x}+{3}}}+\frac{{B}}{{{2}{x}-{3}}}$

Multiply through the the LCD $\displaystyle{4}{{x}}^{{2}}-{9}.$

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

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

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

Equate the coefficient of like terms. Then solve for A and B.

$\displaystyle{0}={2}{A}+{2}{B}$

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

Use the elimination method to solve for A and B. Multiply the first equation by 3

and multiply the second equation by 2.

$\displaystyle{0}={6}{A}+{6}{B}$

$\displaystyle{2}=-{6}{A}+{6}{B}$

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

$\displaystyle{2}={12}{B}$

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

$\displaystyle{0}={2}{A}+{2}{B}$

$\displaystyle{0}={2}{A}+{2}{\left(\frac{{1}}{{6}}\right)}$

$\displaystyle{A}=-\frac{{1}}{{6}}$

$\displaystyle{A}=-\frac{{1}}{{6}},{B}=\frac{{1}}{{6}}$

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}=-\frac{{1}}{{{6}{\left({2}{x}+{3}\right)}}}+\frac{{1}}{{{6}{\left({2}{x}-{3}\right)}}}$

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}=\frac{{1}}{{{\left({2}{x}+{3}\right)}{\left({2}{x}-{3}\right)}}}$

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}=\frac{{A}}{{{2}{x}+{3}}}+\frac{{B}}{{{2}{x}-{3}}}$

Multiply through the the LCD $\displaystyle{4}{{x}}^{{2}}-{9}.$

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

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

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

Equate the coefficient of like terms. Then solve for A and B.

$\displaystyle{0}={2}{A}+{2}{B}$

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

Use the elimination method to solve for A and B. Multiply the first equation by 3

and multiply the second equation by 2.

$\displaystyle{0}={6}{A}+{6}{B}$

$\displaystyle{2}=-{6}{A}+{6}{B}$

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

$\displaystyle{2}={12}{B}$

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

$\displaystyle{0}={2}{A}+{2}{B}$

$\displaystyle{0}={2}{A}+{2}{\left(\frac{{1}}{{6}}\right)}$

$\displaystyle{A}=-\frac{{1}}{{6}}$

$\displaystyle{A}=-\frac{{1}}{{6}},{B}=\frac{{1}}{{6}}$

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}=-\frac{{1}}{{{6}{\left({2}{x}+{3}\right)}}}+\frac{{1}}{{{6}{\left({2}{x}-{3}\right)}}}$

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Monday, November 2, 2015

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}$ 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"?

Monday, November 2, 2015

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;?

$\displaystyle\frac{{1}}{{{4}{{x}}^{{2}}-{9}}}$ 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"?