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Solve ${{\log }_{\dfrac{1}{4}}}(\dfrac{35-{{x}^{2}}}{x})\ge -\dfrac{1}{2}$

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Last updated date: 25th Jul 2024
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Answer
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Hint: If a>$1$ ,
then ${{\log }_{a}}x>{{\log }_{a}}y$
$\Rightarrow $ x > y
and if $0$ < a < $1$
 then ${{\log }_{a}}x>{{\log }_{a}}y$
$\Rightarrow $ x < y.
This means that when we take antilog on both sides of the equation, we have to reverse the inequalities. Also note that
${{\log }_{a}}{{a}^{k}}=k$ ……($2$)
and
${{a}^{(\log {{a}^{k}})}}=k$ ……($3$)
for any number k and a > $0$

Complete step by step solution:
We first raise both sides to the power of $\dfrac{1}{4}$. Then using the Hint we get
$\begin{align}
  & {{\dfrac{1}{4}}^{{{\log }_{\dfrac{1}{4}}}(\dfrac{35-{{x}^{2}}}{x})}}\le {{\dfrac{1}{4}}^{-\dfrac{1}{2}}} \\
 & \Rightarrow (\dfrac{35-{{x}^{2}}}{x})\le {{\dfrac{1}{4}}^{-\dfrac{1}{2}}} \\
 & \Rightarrow (\dfrac{35-{{x}^{2}}}{x})\le {{4}^{\dfrac{1}{2}}} \\
 & \Rightarrow (\dfrac{35-{{x}^{2}}}{x})\le 2 \\
\end{align}$
with the inequality reversed as $0<\dfrac{1}{4}<1$. We have also used the fact that
$\dfrac{1}{{{a}^{k}}}={{a}^{-k}}$ for any numbers $a$ and $k$.
Now the inequality is a simple polynomial inequality and can be solved as
$\begin{align}
  & (\dfrac{35-{{x}^{2}}}{x})\le 2 \\
 & \Rightarrow 35-{{x}^{2}}\le 2x \\
 & \Rightarrow {{x}^{2}}+2x-35\ge 0 \\
\end{align}$
This quadratic equation can be easily factorized which gives
$\begin{align}
  & {{x}^{2}}+(7-5)x-(7\times 5)\ge 0 \\
 & \Rightarrow (x-5)(x+7)\ge 0 \\
\end{align}$
In the last step we only need to realize that this expression is positive only when either of the terms are positive or when both are negative. Since x > $5$ clearly means that x > $-7$ we have one range of solutions as x >$5$ . Also x < $-7$ would also make the first factor negative so we have the other range as x < $-7$. The middle part has the expression negative since the first bracket is negative and second positive.
So we write the solutions as
$x\in (-\infty ,-7]\cup [5,\infty )$

Note:
The reversal of inequality is necessary. A common error would be to not do that and that would lead to the erroneous solution of x between $-7$ and $5$.