Question

solving quadratic equations. z(z-1)=0

Answer 1

If

z(z-1)=0

then

z = 0 or z-1=0

Answer 2

Keep going to completely solve for z

Answer 3

i don't understand how to do the problem? could you explain please? sorry. haven't been in algebra for quite sometime, but the help is much appreciated!

Answer 4

If two numbers multiply to 0, then one of them must be 0

Answer 5

The quadratic EQUATION is
\[solution =\frac{b \pm \sqrt{b^2-4 a c}}{2 a}\]
which holds true for any
\[a z^2 + b z + c = 0\]
Where a, b, and c are real numbers and z is complex (and, therefore, can be real).
You have to get z(z-1) = 0 in the form I showed above to employ the quadratic equation. Then again, if you are given z(z-1) as is, in factored form, then it is easier the other way...

Answer 6

So if

z(z-1)=0

then

either z is zero or z-1 is zero, which means

z=0 or z-1=0

which becomes

z=0 or z=1

Answer 7

oh my goodness, that makes perfect sense now! thank you so much @jim_thompson5910!

Answer 8

you're very welcome

Answer 9

I'm glad it's making sense now

Answer 10

would you mind helping with another problem and explaining as well? if that wouldn't be too much of a bother?

y^2+y-30=0
i've gotten to this far in the equation:
y^2=0 or y-30=0

Answer 11

not quite

Answer 12

you can only say that if y^2 times y-30 was equal to 0

Answer 13

so you have to factor the left side first

Answer 14

Do you know how to factor this?

Answer 15

i'm factoring it right now

Answer 16

well you don't have to factor, if you read what I wrote

Answer 17

@jim_thompson5910 okay, i worked it like this:

y^2+y-30=0
(y+5)(y-6)=0
y+5=0 or y-6=0
y=-5 or y=6

is that correct?

Answer 18

close, but not quite

Answer 19

-6 + 5 = -1, but we want the two numbers to add to positive 1

Answer 20

but you have the right idea (minus that mistake)

Answer 21

so it'd be:

y^2+y-30=0
(y+6)(y-5)=0
y+6=0 or y-5=0
y=-6 or y=5

correct?

Answer 22

much better

Answer 23

thank you! could you help with this problem as well? lol

3c^2=5c

Answer 24

get everything to one side

Answer 25

what do you get when you do that

Answer 26

i simplified it wrong, i got a fraction..

Answer 27

that's normal (and in this case correct)

Answer 28

depends on which fraction of course

Answer 29

5/3 or 1and 2/3

Answer 30

that's one solution

Answer 31

what's the other

Answer 32

2?

Answer 33

@jim_thompson5910 still helpin me?

Answer 34

oh sry, i got distracted

Answer 35

no, 2 isn't the other solution

Answer 36

well darn, idk then

Answer 37

3c^2=5c

3c^2-5c = 0

c(3c - 5) = 0

c = 0 or 3c-5=0

c = 0 or c = 5/3

So the two solutions are c = 0 or c = 5/3

Answer 38

of course, you can write 5/3 as 1 2/3 (a mixed number)

Answer 39

not sure if you did it this way, but this is the way I recommend doing it

Answer 40

yeah i definitely did it some crazy way, but that way makes more sense

Answer 41

alright I'm glad it makes sense

Answer 42

okay what about this problem...

10x^2+9x+2=0
(-5x+2)(-2x+1)=0
-5x+2=0 or -2x+1=0
idk what to do after that^^

Answer 43

unfortunately 10x^2+9x+2 doesn't factor to (-5x+2)(-2x+1)

Answer 44

oh okay.

Answer 45

try it again and tell me what you get

Answer 46

umm i guessed but i got (-5x+3)(3x+1)

Answer 47

still no

10*2 = 20 (first and last multiply to 20)

Notice how 5+4 = 20 and 5*4 = 20


So break up 9x into 5x+4x and factor by grouping


10x^2+9x+2

10x^2+5x+4x+2

(10x^2+5x)+(4x+2)

5x(2x+1)+(4x+2)

5x(2x+1)+2(2x+1)

(5x+2)(2x+1)

So 10x^2+9x+2 factors to (5x+2)(2x+1)

Answer 48

so then it would be 5x+2 =0 or 2x+1=0 right?

Answer 49

yes, keep going

Answer 50

5-2=0 or 2-1=0?

Answer 51

not sure what you're doing there...

Answer 52

wait x=5-2 or x=2-, i messed up

Answer 53

closer, but a bit off

Answer 54

5x+2 =0

5x+2-2 =0-2

5x =-2

5x/5 = -2/5

x = - 2/5

Answer 55

2x+1=0

2x+1-1=0-1

2x = -1

2x/2 = -1/2

x = -1/2

Answer 56

okay thank you

Answer 57

yw