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August 20th, 2017, 05:23 AM  #1 
Newbie Joined: Aug 2017 From: Bath, UK Posts: 5 Thanks: 0  Solve Inequalities
2(x^22)>7x Expand out the brackets: 2x^24>7x Tidy it up a little: 2x^27x>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. Many thanks. 
August 20th, 2017, 06:06 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 785 Thanks: 312 
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. 
August 20th, 2017, 06:19 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,628 Thanks: 1308 
$2(x^22) > 7x$ $2x^2 7x4 > 0$ note that the graph of $y=2x^27x4$ is an upward opening parabola where $y>0$ outside its zeros and $y < 0$ between its zeros. $(2x +1)(x4) > 0$ zeros are $x=\dfrac{1}{2}$ and $x=4$ ... inequality solution set is $\left\{ x \in \left(\infty, \dfrac{1}{2}\right) \cup (4, \infty) \right\}$ 
August 22nd, 2017, 04:07 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,656 Thanks: 681 
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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