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August 20th, 2017, 05:23 AM   #1
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Solve Inequalities

2(x^2-2)>7x
Expand out the brackets:
2x^2-4>7x
Tidy it up a little:
2x^2-7x>4
But where do I go from here? The real question is, how can I go about removing the x^2, I can never quite remember the rules.

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August 20th, 2017, 06:06 AM   #2
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The trick with inequalities is to solve the related equality first. In this case you have to remember the quadratic formula to do that.

$2(x^2 - 2) = 7x \implies 2x^2 - 4 = 7x \implies 2x^2 - 7x - 4 = 0 \implies$

$x = \dfrac{-\ (-\ 7) \pm \sqrt{(-\ 7)^2 - 4(2)(-\ 4)}}{2 * 2} = \dfrac{7 \pm \sqrt{49 +32}}{4} = \dfrac{7 \pm \sqrt{81}}{4} \implies$

$x = 4 \text { or } x = -\ 0.5.$

Now you can address the inequality.

$2x^2 - 7x - 4 > 0 \implies 2(x^2 - 2) > 7x.$

But, from having solved the related equality, you know that

$2\{(-\ 0.5)^2 - 2\} = 2(0.25 - 2) = 2(- 1.75) = -\ 3.5 = 7(-\ 0.5)$ and

$2(4^2 - 2) = 2 * 14 = 7 * 4.$

So test a value less than - 0.5, a value between - 0.5 and 4, and a value above 4.
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August 20th, 2017, 06:19 AM   #3
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$2(x^2-2) > 7x$

$2x^2 -7x-4 > 0$

note that the graph of $y=2x^2-7x-4$ is an upward opening parabola where $y>0$ outside its zeros and $y < 0$ between its zeros.

$(2x +1)(x-4) > 0$

zeros are $x=-\dfrac{1}{2}$ and $x=4$ ...

inequality solution set is $\left\{ x \in \left(-\infty, -\dfrac{1}{2}\right) \cup (4, \infty) \right\}$
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August 22nd, 2017, 04:07 AM   #4
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Equivalently, solve the associated equation first.

To solve $\displaystyle 2x^2- 7x- 4> 0$, first solve $\displaystyle 2x^2- 7x- 4= 0$.

Using the quadratic formula or "completing the square", $\displaystyle x= \frac{7\pm\sqrt{49+ 32}}{4}$$\displaystyle = \frac{7\pm\sqrt{81}}{4}=$$\displaystyle \frac{7\pm 9}{4}$.

So $\displaystyle x= \frac{7+ 9}{4}= 4$ and $\displaystyle x= \frac{7- 9}{4}= -\frac{1}{2}$ make the polynomial equal to 0. Since all polynomials are continuous functions, those are the only places the polynomial can change sign. Those two points divide the number line into 3 intervals, x< -1/2, -1/2< 0< 4, and x> 4. We need only check one point in each interval.

If x= -1< -1/2, $\displaystyle 2x^2- 7x- 4= 2(-1)^2- 7(-1)- 4= 2+ 7- 4= 5> 0$. The inequality is true for all x< -1/2.

If x= 0, -1/2< 0< 4, $\displaystyle 2x^2- 7x- 4= 2(0)^2- 7(0)- 4= -4< 0$. The inequality is not true for any x between -1/2 and 4.

If x= 5> 4, $\displaystyle 2x^2- 7x- 4= 2(5)^2- 7(5)- 4= 50- 35- 4= 11> 0$. The inequality is true for all x> 4.

The original inequality is satisfied for all x< -1/2 and for all x> 4: {x | x< -1/2 or x> 4}.
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