Why Is Pre-Calculus Challenging for Some Students?

[jubej]

hello! I am a studen from sweden and i having some troble solves this kind of problems. i have tried but the book its really bad to expleain. and i really have to know this for the comming exan, so if anyone could help me i really be so happy :)

here are some stuff i have hard time solving:

1) z^4=16i
2) 2z³ - 3z² + 2z -3 = 0


thnx again for any help

 

[rocomath]

hello! I am a studen from sweden and i having some troble solves this kind of problems. i have tried but the book its really bad to expleain. and i really have to know this for the comming exan, so if anyone could help me i really be so happy :)

here are some stuff i have hard time solving:

1) z^4=16i
2) 2z³ - 3z² + 2z -3 = 0


thnx again for any help

are those 2 separate problems or do i solve for 1 and apply it to 2?

1 - clarify question or how do you get an imaginary number?

\sqrt{-X}=\sqrt{X}i

2 - factor out a common term, hint z^2

 

[jubej]

there are 2 separate problems.

the first one its so hard i don't where to start. or how to solve it

and the second one ill start to factor out term z² and see what happens thnx.

 

[jubej]

i can't factor out 2² in 2)

because : 2z³ - 3z² + 2z -3 = 0 has 2z so its not and z² type.
what to do?

 

[rocomath]

not 2^2 but z^2, you eventually get

z^{2}(2z-3)+(2z-3)=0

what would your next step be?

 

[jubej]

that would be maybe

z² = - (2z - 3)/(2z - 3)
z² = -1

its that right?

 

[rocomath]

z^{2}(2z-3)+(2z-3)=0

there is an implied one infront of the 2nd parenthesis

so

z^{2}(2z-3)+1(2z-3)=0

pull out a common term of 2z-3

(z^{2}+1)(2z-3)=0

yes?

 

[HallsofIvy]

Do you know how to put a complex number in "polar form"?

Do you know DeMoivre's theorem? Those are necessary for problem 1.