Find the Set of Values for 'x' and 'z'

[Petrus]

Hello MHB,
I am solving some old exam and I am stuck at one problem

it says 'outline the following two amounts'
If we start with number 1.
we know for the part \(\displaystyle (x-1)(x-5)\leq0\) x has to be \(\displaystyle 1\leq x \leq 5\)
part number 2.
\(\displaystyle (x-3)(x-4)>0\) we know that \(\displaystyle x>4\)
part number 3.
\(\displaystyle (x-2)(x-6)\leq0\) we have\(\displaystyle 2\leq x\leq6\)
for number 1 and 2 we got
\(\displaystyle 4 < x \leq 5\)
and part 2 and 3 we got:
\(\displaystyle 4<x \leq 6\)
Is this correct?

Regards,
\(\displaystyle |\rangle\)

 

[HallsofIvy]

Re: Union

Hello MHB,
I am solving some old exam and I am stuck at one problem
https://www.physicsforums.com/attachments/767
it says 'outline the following two amounts'
If we start with number 1.
we know for the part \(\displaystyle (x-1)(x-5)\leq0\) x has to be \(\displaystyle 1\leq x \leq 5\)
part number 2.
\(\displaystyle (x-3)(x-4)>0\) we know that \(\displaystyle x>4\)
part number 3.
\(\displaystyle (x-2)(x-6)\leq0\) we have\(\displaystyle 2\leq x\leq6\)
for number 1 and 2 we got
\(\displaystyle 4 < x \leq 5\)
and part 2 and 3 we got:
\(\displaystyle 4<x \leq 6\)
Is this correct?

Regards,
\(\displaystyle |\rangle\)

What you have said so far is correct but you have not finished the problem. The intersection is the set of all numbers that are in all of those intervals. You have that some of the numbers are in 4< x< 5 and some are in 4< x< 6. Numbers that are in both are all the numbers in 4< x< 5.

 

[Petrus]

Re: Union

What you have said so far is correct but you have not finished the problem. The intersection is the set of all numbers that are in all of those intervals. You have that some of the numbers are in 4< x< 5 and some are in 4< x< 6. Numbers that are in both are all the numbers in 4< x< 5.

Hello HallsofIvy,
so for all those three range the answer to the question is \(\displaystyle 4< x \leq 5\)?

Regards,
\(\displaystyle |\rangle\)

 

[alyafey22]

Re: Union

Hello HallsofIvy,
so for all those three range the answer to the question is \(\displaystyle 4< x \leq 5\)?

Regards,
\(\displaystyle |\rangle\)

According to your answer the only integer solution is 5 , check if that is true .

 

[Petrus]

Re: Union

According to your answer the only integer solution is 5 , check if that is true .

Hello ZaidAlyafey,
the facit says \(\displaystyle [2,3)\)U\(\displaystyle (4,5]\) and I don-t understand how :S

Regards,
\(\displaystyle |\rangle\)

 

[alyafey22]

Re: Union

part number 2.
\(\displaystyle (x-3)(x-4)>0\) we know that \(\displaystyle x>4\)

Recheck this one .

 

[Petrus]

Re: Union

Recheck this one .

Hello ZaidAlyafey,
Darn it... I see... I did NOT think clear...we got \(\displaystyle x>4\) or \(\displaystyle x<3\)
so we got \(\displaystyle 4<x \leq 5\) or \(\displaystyle 2 \leq x <3\) Thanks alot!:) I will keep try b and will be back if I am still stuck!
Regards,
\(\displaystyle |\rangle\)

 

[Petrus]

Hello,
I am back for the b) This is my progress:
we got \(\displaystyle Re[(z-1)(\overline{z}+i)]=\frac{1}{2}\) so we start with
\(\displaystyle (z-1)(\overline{z}+i) = z\overline{z}+zi-\overline{z}-i\)
\(\displaystyle z=x+yi\)
\(\displaystyle \overline{z}=x-yi\)
we get that:
\(\displaystyle x^2+y^2+xi-y-x+yi-i\)
with other words we got this equation
\(\displaystyle x^2+y^2-y-x=\frac{1}{2}\) (Re)
\(\displaystyle x+y-1=0 <=> x+y=1\) (Im)
Now I am stuck on how to solve them

Regards,
\(\displaystyle |\rangle\)

 

[agentredlum]

Hi Petrus, at the bottom, in your x + y = 1 , that 1 should be i

:)

 

[Petrus]

Hi Petrus, at the bottom, in your x + y = 1 , that 1 should be i

:)

Hello agentmulder,
I did factour out i, that is why I got 1, I should write \(\displaystyle i(x+y-1)=0i\)

Regards
\(\displaystyle |\rangle\)

 

[alyafey22]

\(\displaystyle Re[(z-1)(\overline{z}+i)]=\frac{1}{2}\) so we start with

You cannot conclude that \(\displaystyle \text{Im}[(z-1)(\overline{z}+i)] =0 \)

 

[Petrus]

You cannot conclude that \(\displaystyle \text{Im}[(z-1)(\overline{z}+i)] =0 \)

Ok I thought I could, but how shall I solve that Re equation?

Regard
\(\displaystyle |\rangle\)

 

[agentredlum]

Hello agentmulder,
I did factour out i, that is why I got 1, I should write \(\displaystyle i(x+y-1)=0i\)

Regards
\(\displaystyle |\rangle\)

Oh shoot, missed that, you're right of course.

:)

 

[alyafey22]

\(\displaystyle x^2+y^2-y-x=\frac{1}{2}\) (Re)

First note that the solution will not be a finite number of points since we have an equation with two variables

We complete the square

\(\displaystyle x^2-x+\frac{1}{4}- \frac{1}{4} +y^2- y+ \frac{1}{4}-\frac{1}{4} = \frac{1}{2}\)

\(\displaystyle \left( x-\frac{1}{2} \right)^2+ \left(y-\frac{1}{2} \right)^2=1\)

so what do you conclude ?

 

[Petrus]

First note that the solution will not be a finite number of points since we have an equation with two variables

We complete the square

\(\displaystyle x^2-x+\frac{1}{4}- \frac{1}{4} +y^2- y+ \frac{1}{4}-\frac{1}{4} = \frac{1}{2}\)

\(\displaystyle \left( x-\frac{1}{2} \right)^2+ \left(y-\frac{1}{2} \right)^2=1\)

so what do you conclude ?

Hello Zaid,
I get problem when solving that equation, if \(\displaystyle x,y=\frac{1}{2}\) it will be zero? Well facit says it should be \(\displaystyle x,y=\frac{1}{2}\) but I don't understand

Regards,
\(\displaystyle |\rangle\)

 

[alyafey22]

Hello Zaid,
I get problem when solving that equation, if \(\displaystyle x,y=\frac{1}{2}\) it will be zero? Well facit says it should be \(\displaystyle x,y=\frac{1}{2}\) but I don't understand

Regards,
\(\displaystyle |\rangle\)

I cannot see why it should be just one point , the set of solutions will lie on the circle with center \(\displaystyle \left( {1 \over 2} , {1\over 2} \right) \,\, \) and radius \(\displaystyle r=1\)

 

[Petrus]

I cannot see why it should be just one point , the set of solutions will lie on the circle with center \(\displaystyle \left( {1 \over 2} , {1\over 2} \right) \,\, \) and radius \(\displaystyle r=1\)

Hello Zaid,
I see they did prob mean that :P May I ask you how did you think when you did that factour part, how did you know you would add and take away \(\displaystyle \frac{1}{4}\)? I need to practice those thing, any tips?

Regards,
\(\displaystyle |\rangle\)

 

[alyafey22]

Hello Zaid,
I see they did prob mean that :P May I ask you how did you think when you did that factour part, how did you know you would add and take away \(\displaystyle \frac{1}{4}\)? I need to practice those thing, any tips?

Regards,
\(\displaystyle |\rangle\)

Suppose we have the following \(\displaystyle x^2+bx+c\) and you want to complete the square ,

Find the coefficient of x , which is b .

Divide it by 2 and square the result , \(\displaystyle \frac{b^2}{4}\).

Add and subtract to have a complete square .

\(\displaystyle x^2+bx+\frac{b^2}{4} -\frac{b^2}{4} + c = \left( x+ \frac{b}{2} \right)^2 + c -\frac{b^2}{4} \)

 

[Petrus]

Thanks evryone for all help and taking your time! Now I do understand but practice need :)

Regards,
\(\displaystyle |\rangle\)

 

[alyafey22]

You might find this http://www.mathhelpboards.com/f21/teach-complete-square-3817/beneficial .