Notes: Solve for x. x^2+2x+3/2x-3 - 2x-3/x^2+2x+3 = 35/6Show complete solution and explain the answer

Monday, November 4, 2013

Solve for x. x^2+2x+3/2x-3 - 2x-3/x^2+2x+3 = 35/6Show complete solution and explain the answer

$\displaystyle\frac{{{{x}}^{{2}}+{2}{x}+{3}}}{{{2}{x}-{3}}}-\frac{{{2}{x}-{3}}}{{{{x}}^{{2}}+{2}{x}+{3}}}=\frac{{35}}{{6}}$

Multiply
by LCD =
$\displaystyle{\left({2}{x}-{3}\right)}{\left({{x}}^{{2}}+{2}{x}+{3}\right)}$

$\displaystyle{{\left({{x}}^{{2}}+{2}{x}+{3}\right)}}^{{2}}-{{\left({2}{x}-{3}\right)}}^{{2}}={\left(\frac{{35}}{{6}}\right)}{\left({2}{x}-{3}\right)}{\left({{x}}^{{2}}+{2}{x}+{3}\right)}$

Using
difference of squares we
get

$\displaystyle{\left({\left({{x}}^{{2}}+{2}{x}+{3}\right)}+{\left({2}{x}-{3}\right)}\right)}{\left({\left({{x}}^{{2}}+{2}{x}+{3}\right)}-{\left({2}{x}-{3}\right)}\right)}=\frac{{35}}{{6}}{\left({2}{x}-{3}\right)}{\left({{x}}^{{2}}+{2}{x}+{3}\right)}$

$\displaystyle{6}{\left({{x}}^{{2}}+{4}{x}\right)}{\left({{x}}^{{2}}+{6}\right)}={35}{\left({2}{{x}}^{{3}}+{4}{{x}}^{{2}}+{6}{x}-{3}{{x}}^{{2}}-{6}{x}-{9}\right)}$

Factor

$\displaystyle{6}{x}{\left({x}+{4}\right)}{\left({{x}}^{{2}}+{6}\right)}={35}{\left({2}{{x}}^{{3}}+{{x}}^{{2}}-{9}\right)}$

Multiply
out

$\displaystyle{6}{x}{\left({{x}}^{{3}}+{6}{x}+{4}{{x}}^{{2}}+{24}\right)}={70}{{x}}^{{3}}+{35}{{x}}^{{2}}-{315}$

$\displaystyle{6}{{x}}^{{4}}+{24}{{x}}^{{3}}+{36}{{x}}^{{2}}+{144}{x}={70}{{x}}^{{3}}+{35}{{x}}^{{2}}-{315}$

Puting
in standard
form

$\displaystyle{6}{{x}}^{{4}}-{46}{{x}}^{{3}}+{{x}}^{{2}}+{144}{x}+{315}={0}$

We
can graph to find possible roots: of 3 and 7

We can divide
to find if it is a root

We can divide
$\displaystyle{6}{{x}}^{{4}}-{46}{{x}}^{{3}}+{{x}}^{{2}}+{144}{x}+{315}$ by (x-3) to see if it is a
root.

3)6 -46 1 144 315
18 -84 -249
-315
6 -28 -83 -105
0
$\displaystyle{\left({x}-{3}\right)}{\left({6}{{x}}^{{3}}-{28}{{x}}^{{2}}-{83}{x}-{105}\right)}={0}$

Now we can divide
$\displaystyle{6}{{x}}^{{3}}-{28}{{x}}^{{2}}-{83}{x}-{105}$ by (x-7) to find if we have another
root.

7)6 -28 -83 -105
42 98
105
6 14 15 0

So
$\displaystyle{\left({x}-{3}\right)}{\left({x}-{7}\right)}{\left({6}{{x}}^{{2}}+{14}{x}+{15}\right)}={0}$

We can use the quadratic
formula to find the roots of $\displaystyle{\left({6}{{x}}^{{2}}+{14}{x}+{15}\right)}$ to find the full
solution

x = 3, x = 7 and two imaginary roots $\displaystyle{\left(-\frac{{7}}{{6}}\pm\right.}$
i(sqrt(41)/6))

Notes

Monday, November 4, 2013

Solve for x. x^2+2x+3/2x-3 - 2x-3/x^2+2x+3 = 35/6Show complete solution and explain the answer

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What is the meaning of the 4th stanza of Eliot's Preludes, especially the lines "I am moved by fancies...Infinitely suffering thing".

Monday, November 4, 2013

Solve for x. x^2+2x+3/2x-3 - 2x-3/x^2+2x+3 = 35/6Show complete solution and explain the answer

No comments:

Post a Comment

What is the meaning of the 4th stanza of Eliot&#39;s Preludes, especially the lines &quot;I am moved by fancies...Infinitely suffering thing&quot;.

What is the meaning of the 4th stanza of Eliot's Preludes, especially the lines "I am moved by fancies...Infinitely suffering thing".