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

#### Petrus

##### Well-known member
Hello MHB,
I am solving some old exam and I am stuck at one problem it says 'outline the following two amounts'
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,
$$\displaystyle |\rangle$$

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#### HallsofIvy

##### Well-known member
MHB Math Helper
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'
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
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
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
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
Re: Union

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

#### Petrus

##### Well-known member
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,
$$\displaystyle |\rangle$$

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#### Petrus

##### Well-known member
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
Hi Petrus, at the bottom, in your x + y = 1 , that 1 should be i #### Petrus

##### Well-known member
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
$$\displaystyle Re[(z-1)(\overline{z}+i)]=\frac{1}{2}$$ so we start with
You cannot conclude that $$\displaystyle \text{Im}[(z-1)(\overline{z}+i)] =0$$

#### Petrus

##### Well-known member
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
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
$$\displaystyle x^2+y^2-y-x=\frac{1}{2}$$ (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
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
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
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 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
Hello Zaid,
I see they did prob mean that 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
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
You might find this thread beneficial .