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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^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. 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. Thanks from EvolvedPie
 August 20th, 2017, 06:19 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,628 Thanks: 1308 $2(x^2-2) > 7x$ $2x^2 -7x-4 > 0$ 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 $\left\{ x \in \left(-\infty, -\dfrac{1}{2}\right) \cup (4, \infty) \right\}$
 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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