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I have to teach mathematics to one of the kid in my family. Now, it has been a long time I have not done such exercises so could you help me out ?

The exercise requires to solve the following inequality :

(a) $ |x²-4x| > |x²-4x+a| $

First Method

So one way to solve this is to square both sides. You don't change anything (I believe) to the relationship by squaring it and therefore you can easily obtain the solution by searching the roots of the then obtained equation :

$ 0 > 2ax² - 8ax + a² $

Finding the roots will allow finding the x values which satisfty this inequation. So let's go root finding.

$$ \rho = b^2-4ac = (-8a)^2 - 4 \cdot 2a \cdot a^2 = 64a^2 - 8a^3 = 8a^2(8-a)$$

if $ \rho < 0 $ then inequality can not hold since polynnomial will always be greater than 0. In equation this gives :

$$ 8a^2(8-a) < 0 $$ $$ 8-a < 0 $$ $$ 8 < a $$

if $\rho > 0$ then the roots are the following ones knowing that this implies $a<8$:

$$ \frac{-b \pm \sqrt{\rho}}{2a} = \frac{8a \pm \sqrt{8a^2 \cdot (8-a)}}{4a} = \frac{8a \pm 2a \sqrt{2 \cdot (8-a)}}{4a} = \frac{4 \pm \sqrt{2 \cdot (8-a)}}{2} = 2 \pm \frac{\sqrt{2 \cdot (8-a)}}{2} $$

the inequality is therefore satisfied for $$a < 8 $$ $$ 2 - \frac{\sqrt{2 \cdot (8-a)}}{2} < x < 2 + \frac{\sqrt{2 \cdot (8-a)}}{2} $$

Second method

Ok. Now what if I don't want to use the "trick" of squaring ? Then I can discuss the possibilities depending on the sign of the argument inside the absolute value function.

The following cases are required :

1) $ x²-4x \geq 0$ and $x² -4x +a \geq 0 $

2) $ x²-4x \geq 0$ and $x² -4x +a \leq 0 $

3) $ x²-4x \leq 0$ and $x² -4x +a \geq 0 $

4) $ x²-4x \leq 0$ and $x² -4x +a \leq 0 $

Let's analyze the first case. Then we can write the inequality $(a)$ the following way : $$ x² - 4x > x² -4x+a $$ this provides the condition that : $$ a < 0 $$ The rewritten inequality is obtained only if the following conditions are answered :

(a) $ x²-4x \geq 0 = x(x-4) \geq 0$

and

(b) $x² -4x +a \geq 0 $

The first inequality (a) requires $x\geq 4$ or $ x < 0$. The second inequality (b) requires finding root. This is found by computing $\rho$ :

$$\rho = b^2-4ac = 16-4a $$

If $\rho < 0$ then the second inequality (b) is always true. In other words if $ 4 < a $ then second inequality (b) is always true. However this conflicts the requirement that $a<0$ and is therefore impossible.

If $\rho > 0$ then one finds the roots : $$ 2 \pm \sqrt{4-a} $$

and therefore to answer the inequality (b) this requires : $$ x > 2 + \sqrt{4-a} $$ $$ x < 2 - \sqrt{4-a} $$

Combining all conditions obtained through this analysis : $$ a < 0 $$ $$ x \geq 4 \; or \; x < 0 $$ $$ x > 2 + \sqrt{4-a} $$ $$ x < 2 - \sqrt{4-a} $$

So since $ a < 0$ the square root will always be positive. If $ x < 0$ is true then only second root is applicable. If $ x \geq > 4$ then only first root is applicable. Now this answer already leads to different conditions than the first method. So where do I do something wrong ?

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First of all you have to distinguish four cases: $$x^2-4x\geq 0$$ and $$x^2-4x+a\geq0$$ or $$x^2-4x\geq 0$$ and $$x^2-4x+a<0$$ or $$x^2-4x<0$$ and $$x^2-4x+a\geq 0$$ or $$x^2-4x<0$$ and $$x^2-4x+a<0$$ Can you proceed? I have got $$x<2-\sqrt{4-\frac{a}{2}}$$ and $a<0$ $$x>2+\sqrt{4-\frac{a}{2}}$$ and $a<0$ $$\frac{1}{2}(4-\sqrt{2}\sqrt{8-a})<x<\frac{1}{2}\sqrt{2}\sqrt{8-a}+4)$$ and $0<a<8$

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  • $\begingroup$ Agree with you. As said I only prepared the first case and I find something that does not provide an identical answer to First Method. What you are saying is that maybe by performing the other cases I will find contradictions that in the end will restrict the possibilities to the same answer ? $\endgroup$ – FenryrMKIII Jan 2 at 16:00
  • $\begingroup$ In this case you must solve the inequalities. $\endgroup$ – Dr. Sonnhard Graubner Jan 2 at 16:02
  • $\begingroup$ How do you obtain the $a/2$ inside the square root ? And your answer confirms what I had : you don't get the same conditions with second method compared to first method which is kind of weird no ? $\endgroup$ – FenryrMKIII Jan 2 at 16:27
  • $\begingroup$ If you consider the case $$x^2-4x<0$$ and $$x^2-4x+a\geq 0$$ then you will get $$0>2x^2-8x+a$$ a quadratic inequality to solve. $\endgroup$ – Dr. Sonnhard Graubner Jan 2 at 16:37
  • $\begingroup$ Ok. And why does it seem you obtain nothing from the inequations $$x^2-4x+a \geq 0$$ and $$x^2 -4x \geq 0 $$ $\endgroup$ – FenryrMKIII Jan 2 at 16:43

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