How do I solve this inequality? (1 Viewer)

Bob_man · Dec 25, 2021

Does anyone know how I can solve 5(b)???

=)(= · Dec 25, 2021

From part a where you drew the graphs find where Ix-2I is above 1/x

=)(= · Dec 25, 2021

hope that helps

5uckerberg · Dec 25, 2021

Okay, are you doing Maths extension II? If you do so or not note that the absolute values represent the length of distance from 0 to a certain equation so in this case,
what we have here is that the value has to be either bigger than $\frac{1}{x}$ or smaller than $-\frac{1}{x}$ . With that knowledge in hand, the next step is to state that where is $|x-2|=0$ because this is the point where the line reflects itself off a wall.

Do you see it?

If so congrats because that serves a vital role in what we are doing the reason is that when $x>2$ $|x-2|$ is just simply when is $x-2 > \frac{1}{x}$ and when $x<2$ $|x-2|$ is simply when is $-x+2 < -\frac{1}{x}$ .

Now what I am about to do will attract sceptics but hey the reason why it works will be revealed in a while.

To start off $x-2 > \frac{1}{x}$ . Now meet the critics. "Sir, why are you allowed to say that $x^{2}-2x-1>0$ when normally our maths teacher always tells us to multiply by an even power before doing this?"

The reason is that at $x > 2$ all the values that come after it are positive as thus, it is safe to multiply by x because multiplying by a positive value does not change the result in a fraction. The next step is to find the roots in $x^{2}-2x-1>0$ . If you have found the roots a little challenge awaits you find the root that is greater than 2. If you have done than Horray you are halfway done.

The next step is this find where $-x+2 < -\frac{1}{x}$ . The next step wiill be $-x^{3}+2x^{2} < -x$ . Bring everything to one side and it becomes $-x^{3}+2x^{2}+x < 0, -x(x^{2}-2x-1) < 0$ have a look at this $(x^{2}-2x-1) > 0$ when we went from $-x(x^{2}-2x-1) < 0$ to $x(x^{2}-2x-1) > 0$ is simply $(x+1-\sqrt{2})(x+1+\sqrt{2})$ . The rest is history. There you can find the values that satisfy $|x-2| > \frac{1}{x}$ . If you had to do this question by hand my method will work.

=)(= · Dec 25, 2021

ohhhhh i went the lazy route hahaha

=)(= · Dec 25, 2021

Just clearing some things up so x>2 since the inequality wouldn't work otherwise? And from that you got x^2-2x-1>0 which you solved by completing the square giving x>1 ± √ 2 and since x is positive it is 1+ √ 2?

5uckerberg · Dec 25, 2021

=)(= said:
Just clearing some things up so x>2 since the inequality wouldn't work otherwise? And from that you got x^2-2x-1>0 which you solved by completing the square giving x>1 ± √ 2 and since x is positive it is 1+ √ 2?

The absolute value plays an important role.

5uckerberg · Dec 25, 2021

=)(= said:
View attachment 34513
hope that helps

5uckerberg said:
Okay, are you doing Maths extension II? If you do so or not note that the absolute values represent the length of distance from 0 to a certain equation so in this case,
what we have here is that the value has to be either bigger than $\frac{1}{x}$ or smaller than $-\frac{1}{x}$ . With that knowledge in hand, the next step is to state that where is $|x-2|=0$ because this is the point where the line reflects itself off a wall.

Do you see it?

If so congrats because that serves a vital role in what we are doing the reason is that when $x>2$ $|x-2|$ is just simply when is $x-2 > \frac{1}{x}$ and when $x<2$ $|x-2|$ is simply when is $-x+2 < -\frac{1}{x}$ .

Now what I am about to do will attract sceptics but hey the reason why it works will be revealed in a while.

To start off $x-2 > \frac{1}{x}$ . Now meet the critics. "Sir, why are you allowed to say that $x^{2}-2x-1>0$ when normally our maths teacher always tells us to multiply by an even power before doing this?"

The reason is that at $x > 2$ all the values that come after it are positive as thus, it is safe to multiply by x because multiplying by a positive value does not change the result in a fraction. The next step is to find the roots in $x^{2}-2x-1>0$ . If you have found the roots a little challenge awaits you find the root that is greater than 2. If you have done than Horray you are halfway done.

The next step is this find where $-x+2 < -\frac{1}{x}$ . The next step will be $-x^{3}+2x^{2} < -x$ . Bring everything to one side and it becomes $-x^{3}+2x^{2}+x < 0, x(-x^{2}+2x+1) < 0$ have a look at this $(-x^{2}+2x+1) < 0$ is simply $(x+1-\sqrt{2})(x+1+\sqrt{2})$ . The rest is history. There you can find the values that satisfy $|x-2| > \frac{1}{x}$ . If you had to do this question by hand my method will work.

Have a look at your graph and also the part where I got $x(x^{2}-2x-1) > 0$ . There you can focus on all the values of x that give a positive value

=)(= · Dec 25, 2021

5uckerberg said:
Have a look at your graph and also the part where I got $x(x^{2}-2x-1) > 0$ . There you can focus on all the values of x that give a positive value

so does that mean x>1+√ 2 is the answer?

5uckerberg · Dec 25, 2021

=)(= said:
so does that mean x>1+√ 2 is the answer?

One of them but also note there is another solution. Notice when I said the absolute values it means that you can have $-x+2 > \frac{1}{x}$ when x is less than 2 so therefore what happens there is multiply by $x^{2}$ giving us $-x^{3}+2x^{2}-x > 0$ factorising gives us $-x(x^{2}-2x+1) > 0$ changing the sign means changing the direction giving us $x(x^{2}-2x+1) < 0$ now can you observe what is going on in $x^{2}-2x+1$ that equation will always be greater or equal to zero which means he is irrelevant in the problem and then all you got to do is ask yourself what else can I do so that $x(x^{2}-2x+1) < 0$ is true. The answer suddenly is very clear and you have $x < 0$ and combining that with what you said we have $x > 1+\sqrt{2}$ and $x < 0$ .

Two conditions provided.

Apologies for not clarifying correctly and I wish you a Merry Christmas.

=)(= · Dec 25, 2021

ohh makes sense, thank you very much, have a Merry Christmas and a Happy New Year

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