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I saw this problem online today:

$$ \left|5x-3k\right|\geq3\left|x+4k\right| $$

and attempted to solve it by squaring both sides:

$$ \begin{align} \left(5x-3k\right)^2 &\geq3\left(x+4k\right)^2 \\ 25x^2-30kx+9k^2 &\geq 3x^2+24kx+48k^2\\ 22x^2-54kx-39k^2 &\geq 0 \end{align} $$

But this gives me the disgusting critical points

$$ \frac{54k\pm\sqrt{\left(54k\right)^2+4\cdot22\cdot39k^2}}{2\cdot22} $$

I am wondering, what is the correct way to solve this inequality?

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    $\begingroup$ In the first step, we have $$(3(x+4k))^2\\=3^2(x+4k)^2\\=9(x^2+8kx+16k^2)\\=9x^2+72kx+144k^2$$ $\endgroup$
    – lioness99a
    Jun 14, 2017 at 11:38
  • $\begingroup$ This is not a linear algebra problem. $\endgroup$
    – CY Aries
    Jun 14, 2017 at 13:47

4 Answers 4

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You did not square correctly in your first step. $3^2$ becomes $9$ not $3$.

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You made several mistakes in both sides of your inequality when squaring: \begin{align}(5x-3k)^2&\geq \color{red}(3(x+4k)\color{red})^2\\ 25x^2-30k\color{red}x+9k^2&\geq3^\color{red}2(x+4k)^2\\ 25x^2-30k\color{red}x+9k^2&\geq\color{red}9(x^2+8k\color{red}x+16k^2)\\ 25x^2-30k\color{red}x+9k^2&\geq\color{red}9x^2+\color{red}{72}k\color{red}x+\color{red}{144}k^2\\ \color{red}{16}x^2-\color{red}{102}k\color{red}x-\color{red}{135}k^2&\geq 0\end{align}

Therefore, our critical points become \begin{align}x&=\frac{-(-102k)\pm\sqrt{(-102k)^2-4\times 16\times (-135k^2)}}{2\times 16}\\ &=\frac{102k\pm\sqrt{102^2k^2+8640k^2}}{32}\\ &=\frac{102k\pm\sqrt{19044k^2}}{32}\\ &=\frac{102k\pm\sqrt{19044}\sqrt{k^2}}{32}\\ &=\frac{102k\pm138k}{32}\end{align}

And so, \begin{align}x&=\frac{102k+138k}{32}\\ &=\frac{15}{2}k\\ &\text{or}\\ x&=\frac{102k-138k}{32}\\ &=-\frac 98 k\end{align}

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$$ \begin{align} \left(5x-3k\right)^2 &\geq 9\left(x+4k\right)^2 \\ 25x^2-30kx+9k^2 &\geq 9x^2+72kx+144k^2\\ 16x^2-102kx-135k^2 &\geq 0 \end{align} $$ critical points $$x = \frac{102k\pm \sqrt{19044k^2}}{32} = \frac{102\pm 138k}{32}$$ So $x = \frac{15}{2}k$ or $x = -\frac{9}{8}k$

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    $\begingroup$ You seem to be missing some values under your square root $\endgroup$
    – lioness99a
    Jun 14, 2017 at 11:58
  • $\begingroup$ You've still missed out a $k$ before the $\pm$ sign $\endgroup$
    – lioness99a
    Jun 14, 2017 at 12:02
  • $\begingroup$ @lioness99a You are right. I edited the post. $\endgroup$
    – Ahmed
    Jun 14, 2017 at 13:09
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Method I

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$ (where $a>0$, $\alpha>\beta$), then the solution of the inequality $ax^2+bx+c\ge0$ is $x\le\beta$ or $x\ge \alpha$.

\begin{align} \left|5x-3k\right|&\geq3\left|x+4k\right|\\ (5x-3k)^2&\geq [3(x+4k)]^2\\ (5x-3k)^2- (3x+12k)^2&\geq0\\ [(5x-3k)+(3x+12k)][(5x-3k)-(3x+12k)]&\geq0\\ (8x+9k)(2x-15k)&\geq0\\ \end{align}

When $k\ge0$, $\displaystyle \frac{15k}{2}$ is the larger root and $\displaystyle \frac{-9k}{8}$ is the smaller root of the corresponding quadratic equation. So, the solution to the inequality is $$x\le\frac{-9k}{8}\quad\text{or}\quad x\ge\frac{15k}{2}$$

When $k<0$, $\displaystyle \frac{15k}{2}$ is the smaller root and $\displaystyle \frac{-9k}{8}$ is the larger root of the corresponding quadratic equation. So, the solution to the inequality is $$x\le\frac{15k}{2}\quad\text{or}\quad x\ge\frac{-9k}{8}$$


Method II

Let $r\ge0$. Then $|y|\ge r$ if and only if $y\le -r$ or $y\ge r$.

Case (1) If $x\ge -4k$, then $|x+4k|=x+4k$

\begin{align} \left|5x-3k\right|&\geq3\left|x+4k\right|\\ 5x-3k\le-3x-12k\quad&\text{or}\quad 5x-3k\ge3x+12k\\ x\le\frac{-9k}{8}\quad&\text{or}\quad x\ge\frac{15k}{2}\\ \end{align}

If $k\ge0$, $\displaystyle \begin{cases}x\ge -4k \\ \displaystyle x\le\frac{-9k}{8}\quad\text{or}\quad x\ge\frac{15k}{2} \end{cases}$ is equivalent to $-4k\le \displaystyle x\le\frac{-9k}{8}$ or $\displaystyle x\ge\frac{15k}{2}$.

If $k<0$, $\displaystyle x\le\frac{-9k}{8}$ or $\displaystyle x\ge\frac{15k}{2}$ implies that $x$ can be any real number. So, $\displaystyle \begin{cases}x\ge -4k \\ \displaystyle x\le\frac{-9k}{8}\quad\text{or}\quad x\ge\frac{15k}{2} \end{cases}$ is equivalent to $x\ge-4k$.

Case (2) If $x< -4k$, then $|x+4k|=-x-4k$

\begin{align} \left|5x-3k\right|&\geq3\left|x+4k\right|\\ 5x-3k\le3x+12k\quad&\text{or}\quad 5x-3k\ge-3x-12k\\ x\le\frac{15k}{2}\quad&\text{or}\quad x\ge\frac{-9k}{8}\\ \end{align}

If $k\ge 0$, $\displaystyle x\le\frac{15k}{2}$ or $\displaystyle x\ge\frac{-9k}{8}$ implies that $x$ can be any real number. So, $\displaystyle \begin{cases}x< -4k \\ \displaystyle x\le\frac{15k}{2}\quad\text{or}\quad x\ge\frac{-9k}{8} \end{cases}$ is equivalent to $x<-4k$.

If $k< 0$, $\displaystyle \begin{cases}x< -4k \\ \displaystyle x\le\frac{15k}{2}\quad\text{or}\quad x\ge\frac{-9k}{8} \end{cases}$ is equivalent to $\displaystyle x\le\frac{15k}{2}$ or $\displaystyle \frac{-9k}{8}\le x<-4k$.

Combining the above results, when $k\ge0$, the solution to the inequality is

$$x\le\frac{-9k}{8}\quad\text{or}\quad x\ge\frac{15k}{2}$$

When $k\ge0$, the solution to the inequality is

$$x\le\frac{15k}{2}\quad\text{or}\quad x\ge\frac{-9k}{8}$$

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