Solve x^5 + 7x^3 + 6x

$\displaystyle{{x}}^{{5}}+{7}{{x}}^{{3}}+{6}{x}<{5}{{x}}^{{4}}+{7}{{x}}^{{2}}+$
2

$\displaystyle{{x}}^{{5}}-{5}{{x}}^{{4}}+{7}{{x}}^{{3}}-{7}{{x}}^{{2}}+{6}{x}-{2}<$
0

Let us solve the
equation:

$\displaystyle{{x}}^{{5}}-{5}{{x}}^{{4}}+{7}{{x}}^{{3}}-{7}{{x}}^{{2}}+{6}{x}-{2}=$
0

We will substitute with x=
1

==> 1 - 5 + 7 - 7 + 6 -2 =
0

==> 2 - 2 =
0

==> 0 = 0

Then x= 1
is a root for the equation.

==> (x-1) is a
factor:

$\displaystyle{{x}}^{{5}}-{5}{{x}}^{{4}}+{7}{{x}}^{{3}}-{7}{{x}}^{{2}}+{6}{x}-{2}={\left({x}-{1}\right)}{\left({{x}}^{{4}}-{4}{{x}}^{{3}}\right.}$
+3x^2-4x +2)

Now we will factor the last
terms.

First we will rearrange
terms.

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

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

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

Now we will factor
x^2+1

$\displaystyle=\Rightarrow{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}-{4}{x}\right)}={\left({{x}}^{{2}}\right.}$
+1)(x^2-4x+2)

Now we will rewrite the
inequality:

$\displaystyle=\Rightarrow{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}-{4}{x}+{2}\right)}<$
0

We know that $\displaystyle{{x}}^{{2}}+{1}$ is always greater than
zero.

$\displaystyle=\Rightarrow{\left({x}-{1}\right)}{\left({{x}}^{{2}}-{4}{x}+{2}\right)}<$
0

Now we will find the
roots.

==> x1=
1

==> $\displaystyle{x}{2}=\frac{{{4}+\sqrt{{{16}-{8}}}}}{{2}}=\frac{{{4}+\sqrt{{{8}}}}}{{2}}=$
2+sqrt2

$\displaystyle=\Rightarrow{x}{3}={2}-$
sqrt2

==> Now we have the
intervals:

$\displaystyle{\left(-\infty,{1}\right)},{\left({1},{2}-\sqrt{{2}}\right)},{\left({2}-\sqrt{{2}},{2}+\sqrt{{2}}\right)},$
and (2+sqrt2, oo)

$\displaystyle{\left(-\infty,{1}\right)}=\Rightarrow$
+

$\displaystyle{\left({1},{2}-\sqrt{{2}}\right)}=\Rightarrow$
-

$\displaystyle{\left({2},{2}-\sqrt{{2}},{2}+\sqrt{{2}}\right)}=\Rightarrow$
+

$\displaystyle{\left({2}+\sqrt{{2}},\infty\right)}=\Rightarrow$
-

Then, solution is:

x $\displaystyle\in{(}$
1, 2-sqrt2) U (2+sqrt2,
oo)

Help Notes

Saturday, March 12, 2016

Solve x^5 + 7x^3 + 6x

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Film: 'Crocodile Dundee' directed by Peter FaimanHow are stereotypical roles upheld and challenged?

Saturday, March 12, 2016

Solve x^5 + 7x^3 + 6x

No comments:

Post a Comment

Film: &#39;Crocodile Dundee&#39; directed by Peter FaimanHow are stereotypical roles upheld and challenged?

Film: 'Crocodile Dundee' directed by Peter FaimanHow are stereotypical roles upheld and challenged?