Homework Clinic

Mathematics Clinic => Grade 9 Mathematics => Topic started by: aero on Oct 4, 2013

Title: How would you solve this mathematical problem?
Post by: aero on Oct 4, 2013

Solve by using the quadratic formula. Give your answers in simplest radical form. Give both real and imaginary roots.

4/z = 3z/z-3

I have a lot of problems that are just like the one about but I just don't know how to go about solving them.

Title: How would you solve this mathematical problem?
Post by: go.lag on Oct 4, 2013

4/z = 0
No solution

Title: How would you solve this mathematical problem?
Post by: ricki on Oct 4, 2013

You need to use the quadratic formula, so you need to get a quadratic function out of what you're given.

4 = 3z^2/(z-3)

4(z-3) = 3z^2

4z-12=3z^2

Then rearrange by subtracting 4z and adding 12, which leaves you:

3z^2-4z+12=0

Then plug into the quadratic formula, with 3=a, -4=b, and 12=c.

I included a link for the quadratic formula in case you don't know it.

Best of luck! :D

Title: How would you solve this mathematical problem?
Post by: ricki on Oct 4, 2013

If {4/z = 3z/z-3} is equal to {4/z = (3z/z)-3} = 3-3 = 0 then no solution.
If {4/z = 3z/z-3} is equal to {4/z = 3z/(z-3) then  z = 0.66 +/-j1.89  where j^2 = -1

I believe you mean by:
4/z = 3z/(z-3) instead of  4/z = 3z/z-3
If so, then
work;
4/z = 3z/(z-3) or
3z^2 -4z+12 = 0
comparing to the eq. ax^2+bx+c or x = {-b+/-sqrt(b^2-4ac} / 2a.  we get:
z = {4 +/-sqrt(16-144)} / 6
= {4 +/-sqrt(-128)}/6 = 4/6 +/-(j11.31)/6....:where j^2 = -1
z = 0.66 +/-j1.89