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

Petrus

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 
Last edited by a moderator:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Re: Union

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

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

Petrus

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 
Last edited by a moderator:

Petrus

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 

agentmulder

Active member
Feb 9, 2012
33
Hi Petrus, at the bottom, in your x + y = 1 , that 1 should be i

:)
 

Petrus

Well-known member
Feb 21, 2013
739
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
\(\displaystyle |\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle Re[(z-1)(\overline{z}+i)]=\frac{1}{2}\) so we start with
You cannot conclude that \(\displaystyle \text{Im}[(z-1)(\overline{z}+i)] =0 \)
 

Petrus

Well-known member
Feb 21, 2013
739
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
\(\displaystyle |\rangle\)
 

agentmulder

Active member
Feb 9, 2012
33
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.

:)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle x^2+y^2-y-x=\frac{1}{2}\) (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

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Well-known member
Feb 21, 2013
739
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,
\(\displaystyle |\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Well-known member
Feb 21, 2013
739
Thanks evryone for all help and taking your time! Now I do understand but practice need :)

Regards,
\(\displaystyle |\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You might find this thread beneficial .