Question

14. For questions 14–19, what are the solutions of the equation?

z2 – 10z + 24 = 0 (1 point)
–4, –6
4, –6
4, 6
–4, 6
15. 2z2 – 4z + 2 = 0 (1 point)
2, –1
, 1
, –1
2, 1

Answer 1

For 14 just factor it. (z-6)(z-4) because both add up to -10 and multiply to +24. Now just set each equal to 0 and solve.
z-6=0
z-4=0

Answer 2

um C ?

Answer 3

It would help if they were labeled but yes.
Another method would be to use the quadratic formula, or graph it and find where it intersects the x-axis.

Answer 4

thanks(: do you know 15 ?

Answer 5

Same concept it just involves an extra step.
So first we factor out a 2:
\[2(z^2-2z+1)\]
now set it equal to zero and divide by the 2, and since anything divided by 0 is 0 it would change anything:
\[z^2-2z+1=0\]
Now just factor and solve:
\[(z-1)(z-1)\]
And again both add up to a -2 and multiply to a +1.

Answer 6

*wouldn't change anything

Answer 7

Gotcha so its either 1,1 or 1,-1

Answer 8

It's 1, 1 because there are no z+1 in the factored equation

Answer 9

ok gotcha(: Can you help me with a few more please ?

Answer 10

Sure.

Answer 11

thanks (:

Answer 12

6. c2 – 10c = 0
a. 0, 10
b .0, over 10
c. 0, –10
d. 1, –10

15 = 8x2 – 14x
a. -4, 3/8
b. -2/5, 4/3
c. -3, 5/8
d. -3/4, 5/2

Answer 13

6x2 – 17x + 13 = 20x2 – 32
a. 5/3, 9/2
b. -5/2, 9/7
c. -5/2, -9/7
d. -5/3, 9/2

20x2 + 7x – 3 = 0
a. -3/5, 1/4
b. 3/5, 1/4
c. -3/5, -1/4
d. 3.5, -1/4

Answer 14

Well let's start with 6. What can we do first, any ideas?

Answer 15

It's like the previous problem except it's just a variable.

Answer 16

ummm

Answer 17

:/
Factor out a c:
\[c(c-10)=0\]
Now separate them and solve:
\[c=0\]
\[c-10=0\]

Answer 18

im sorry lol i suck at alg ! :(

okay hold on

Answer 19

im going to go with B ?

Answer 20

it's a :/

Answer 21

hey hey i was close lol (:

Answer 22

Yup
Okay next one:
15=8x^2-14

Answer 23

worked it out of paper i think its C or D

Answer 24

How did you start?

Answer 25

started with this 8x^2-14

Answer 26

You always start by moving everything to one side so it can equal 0.

Answer 27

so it should be \[8x^2-14x-15\]

Answer 28

yes ^

Answer 29

Now, since this is almost impossible to factor mentally, we have use the quadratic formula \[\frac{ -b+/-\sqrt{b^2-4ac} }{ 2a }\]

Answer 30

Gotha , what do we start with first ?

Answer 31

We just plug in numbers and start solving it. In this equation the a=8, b=-14 and c =-15

Answer 32

im confuesed

Answer 33

it would look like this:
\[\frac{ 14+\sqrt{-14^2-4(8)(-15)} }{ 2(8) } ; \frac{ 14-\sqrt{-14^2-4(8)(-15)} }{ 2(8) }\]

Answer 34

What are you confused about?

Answer 35

i got it look

Answer 36

15-15=8x^2-14x-15
0=8x^2-14x-15
0=(4x+3)(2x-5)
4x+3-3=0-3
4x/4=-3/4
x=-3/4

Answer 37

2x-5+5=0+5
2x/2=5/2
x=5/2

Answer 38

Oh, haha I guess you did. Sorry I made it a bit harder than it needed to be. Just forget what I said earlier :P

Answer 39

lol its fine , skip all those problems bc i got them.

Answer 40

What are the solutions of the equation 2x2 = 2? Use a graph of the related function

Answer 41

Well put it in standard form.

Answer 42

Aren't equation 3 and 1 the same?

Answer 43

sorry ,

Answer 44

2x2 = 2
4 = 2

Answer 45

wait wait oppss i did it wrong!

Answer 46

2x2=8

2x2−8=0

y=2x2−8

Answer 47

Where did that 8 come from..?

Answer 48

bc 2 & 2 ? or the 4 should b in the 8's place

Answer 49

:/

Answer 50

lol dont gimme that face!

Answer 51

Ha, sorry I just that the original equation was 2x^2=2

Answer 52

and you get 1

Answer 53

well rad 1 but yea

Answer 54

-1 , 1 C

Answer 55

equation ?

Answer 56

3?

Answer 57

yes

Answer 58

Yup

Answer 59

Graph the function. Identify the vertex and axis of symmetry.

f(x) = 4x2 –4x – 2

Answer 60

The vertex is -b/2a