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April 12th, 2014, 09:19 PM   #1
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Another inequality

To solve the inequality $\displaystyle x^4+log_23^{(x^2+4)}+5\leq0$.
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April 12th, 2014, 10:27 PM   #2
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\begin{align*}
x^4+log_23^{(x^2+4)}+5 &\leq 0 \\
x^4+\left( x^2 + 4 \right) \frac{\log{3}}{\log{2}} + 5 &\leq 0 \\
x^4 + \frac{\log{3}}{\log{2}} x^2 + \left(\frac{4\log{3}}{\log{2}} + 5\right) &\leq 0 \\
x^2 = \frac{1}{2}\left( -\frac{\log{3}}{\log{2}} \pm \sqrt{ \frac{\log^2{3}}{\log^2{2}} - 4\left(\frac{4\log{3}}{\log{2}} + 5\right) } \right)
\end{align*}

I reckon we have no real roots here, so I suppose that this is another one where we want to make the imaginary part of the equation disappear for complex $x$.

Thus $x = a +b\imath$

\begin{align*}
\operatorname{Im}{\left(x^4 + \frac{\log{3}}{\log{2}} x^2\right)} &= 0 \\
\operatorname{Im}{\left(a^4 + 4a^3b\imath - 6a^2b^2 -4ab^3\imath + b^4 + \frac{\log{3}}{\log{2}} \left(a^2 + 2ab\imath - b^2 \right) \right)} &= 0 \\
4a^3b -4ab^3 + \frac{\log{3}}{\log{2}} 2ab &= 0 \\
\end{align*}

Which gives $b = 0$ (which we have already dismissed) or
$$ b^2 = a^2 + \frac{\log{3}}{2\log{2}} $$
Which we substitute into
\begin{align*}
\operatorname{Re}{\left(a^4 + 4a^3b\imath - 6a^2b^2 -4ab^3\imath + b^4 + \frac{\log{3}}{\log{2}} \left(a^2 + 2ab\imath - b^2 \right) \right)} &\leq -\left(\frac{4\log{3}}{\log{2}} + 5\right) \\
a^4 - 6a^2b^2 + b^4 + \frac{\log{3}}{\log{2}} \left(a^2 - b^2 \right) &\leq -\left(\frac{4\log{3}}{\log{2}} + 5\right) \\
a^4 - 6a^2\left(a^2 + \frac{\log{3}}{2\log{2}}\right) + \left(a^2 + \frac{\log{3}}{2\log{2}}\right)^2 + \frac{\log{3}}{\log{2}} \left(a^2 - \left(a^2 + \frac{\log{3}}{2\log{2}}\right) \right) &\leq -\left(\frac{4\log{3}}{\log{2}} + 5\right) \\
-4a^4 -\frac{2\log{3}}{\log{2}}a^2 - \frac{\log^2{3}}{4\log^2{2}} &\leq -\left(\frac{4\log{3}}{\log{2}} + 5\right) \\
4a^4 + \frac{2\log{3}}{\log{2}}a^2 + \frac{\log^2{3}}{4\log^2{2}} -\frac{4\log{3}}{\log{2}} - 5 &\ge 0 \\
\end{align*}

This we can solve for a^2, but it's half-past one in the morning, I'm going to bed instead.

I'd be very surprised if that algebra is without errors.
Thanks from Olinguito

Last edited by v8archie; April 12th, 2014 at 10:39 PM.
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April 13th, 2014, 05:55 AM   #3
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Quote:
Originally Posted by v8archie View Post
This we can solve for a^2, but it's half-past one in the morning, I'm going to bed instead.

I'd be very surprised if that algebra is without errors.
I do not understand!For what values of $\displaystyle x$ is satisfied the inequality?

Last edited by Dacu; April 13th, 2014 at 06:28 AM.
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April 13th, 2014, 06:40 PM   #4
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I didn't go as far as getting the values of $x$ because the algebra is getting far too horrible.

Suffice to say that $x$ will be complex, satisfying $x = a + b\imath$ where $a$ and $b$ satisfy equations given in my post above.

If my efforts are correct, the question is phrased badly and should say something like "Find values of $z \in \mathbb{C}$ such that $z^4+log_23^{(z^2+4)}+5$ is real and less than or equal to zero."
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April 13th, 2014, 08:47 PM   #5
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Quote:
Originally Posted by v8archie View Post
I didn't go as far as getting the values of $x$ because the algebra is getting far too horrible.

Suffice to say that $x$ will be complex, satisfying $x = a + b\imath$ where $a$ and $b$ satisfy equations given in my post above.

If my efforts are correct, the question is phrased badly and should say something like "Find values of $z \in \mathbb{C}$ such that $z^4+log_23^{(z^2+4)}+5$ is real and less than or equal to zero."
Any equation of any degree $\displaystyle a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0=0$ can have real solutions and/or complex and so and any inequality of this type,because any inequality can turn into an equation.
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