Question

anyone good with algebra 2? i need help..

Answer 1

Do you have a particular question we can help you with? :)

Answer 2

yea im having a hard time understanding quadratic equations

Answer 3

Anything more specific?

A general quadratic equation is in the form, \(ax^2 + bx + c = 0\). What are you currently doing with these; factoring, solving, or completing squares? Or something more?

Answer 4

uhm here lemme show you what im doing

Answer 5

i just wanna know how to do the warm upsa

Answer 6

Are these supposed to be written that way?
x^2 - 4x - 5 = 0x^2 - 4x - 5 = 0
8x^2 + 40x + 50 = 08x^2 + 40x + 50 = 0

Answer 7

yea but they are 5 different questions

Answer 8

It is confusing that we have two equality signs per question, and some don't even equal the same thing. I can't decipher what the purpose of it...

Answer 9

the first question is x^2-4x-5=0
and i dont know what to do

Answer 10

So the first step says to identify the a, b, and c values.
That is coming from: \(ax^2 + bx + c = 0 \)
\(\color{red}ax^2 + \color{green}bx + \color{blue}c = 0 \)
\(\color{red}1x^2 \color{green}{- 4}x \color{blue}{- 5} = 0 \)
Can you see how these compare so closely?

Answer 11

If there is no coefficient, we can assume it is 1. So for that x^2, it is equally expressed as 1x^2.

Answer 12

So for the first one, can you see from how these two equations compare that a=1, b=-4, and c=-5? I will give this one for a good reference point because there are others for you to try on your own as well.

Answer 13

i see that, and is there a way to solve it or does it just stay 1x2−4x−5=0

Answer 14

We could technically solve for x if we needed (factoring looks applicable, also quadratic formula), but the questions do not ask us for the actual solutions! So no need to do the extra work!

Answer 15

Awesome Sauce! thank you so much ^-^

Answer 16

We were asked to complete the following:
(a) Identify a, b, and c. Which we did for the first one.
(b) Find the discriminant, or in other words take the a, b, and c values from above and apply them to a formula: b^2 - 4ac.
(c) Use that value from (b) to decide the number of solutions.

The discriminant comes from the Quadratic Formula:
\( x = \dfrac{ -b \pm \sqrt{ \color{goldenrod}{b^2 - 4ac}}}{2a} \leftarrow There\ it\ is\)
The square root argument decides the number and type of solutions. If that discriminant is positive, we have -b + sqrt(#) and -b - sqrt(#). Two real solutions. If the discriminant is 0, the square root is 0; so we have -b + or - 0, or just -b. One real solution.
If it is negative discriminant, we have to break out the complex numbers and we get two complex solutions!

Answer 17

so complex is imaginary?
so my answer for number one would be
a- 1x2−4x−5=0
b- x= \[-b \pm \sqrt{b ^{2}}-4ac/2a\]
c- two real solutions ?

Answer 18

Yeah, for the most part. Technically, complex is a better term than imaginary because complex means a real number added to an imaginary number, which is strictly something with i. But if they don't worry about that detail you shouldn't need to either.


For (a) we specifically list our values of a, b, and c. Like, the equation is what they gave us, but we want to provide what is called a, b, or c? a=1, b=-4, or c=-5.
So our answer would literally be just that: (a) a=value, b=value, and c=value.

For (b) the values from (a) are used in the equation b^2 - 4ac. From above, we had a=1, b=-4, and c=-5. Plug those in:
Our answer would be something like...
Discriminant: b^2 - 4ac = (-4)^2 - 4*(1)*(-5) = simplify this to get the answer.

Answer 19

so is the discriminant -4?

Answer 20

Not quite, the subtraction and minus sign combine to make a positive, right? :)

Answer 21

Righto oops. i have been cramming all day because i have to do like 200 something classes before june 13th so im frettin. so it would be 26?

Answer 22

16 + 20 = 36, which I think you meant to put. That is correct.

Answer 23

yea cx thank you so much this helped so much ^-^

Answer 24

Glad to help! And good luck on all those classes! That seems difficult to keep up with. o.o

Answer 25

thank you haha ima need it. yea its mainly because i switched over cyber schools, and the one school i failed with a 64 in chem and a 63 in algebra 2 for the first semester, but in this school passing is a 62. and the school wouldnt give me passing for them so now i have to do both 1st and 2nd semesters for those two classes plus history, english, and some drawing classes.

Answer 26

Ugh, yea, very difficult indeed! Well, I hope you can get through it! I'll be here at least a bit every day if you wanted to tag me for a question (@AccessDenied), although sometimes I just leave the site open while I am gone. But I will try to help you when you need it! And I hope others will also do the same. :D

Answer 27

thank you so much. im gonna go try the second one and come back to you and show you what i got and se if i got it right.

Answer 28

for the second one we use the b^2-4ac and we plug in the stuff from a?

Answer 29

Yep, we will always take those values of a, b, and c for use in the formula. b^2 - 4ac.

Answer 30

okay let me go through this and see what i get ^.^

Answer 31

mehh i think i did somethink wrong Dx

Answer 32

i keep getting this b^2-4ac=(-4)^2-4*(8)*(50)= -1592

Answer 33

i dont know if -1592 is right or?

Answer 34

The original equation is: 8x^2 - 4x + 50 = 0 ?

Answer 35

yeap

Answer 36

Not 8x^2 + 40x + 50 = 0 ? I thought I read that on the sheet earlier.

Answer 37

Oh it is 40 xD my bad again long day lemme retry this

Answer 38

now it comes out to 0 .-.
is that right?

Answer 39

That looks better!

Answer 40

so itd be one real solution correct?

Answer 41

That is correct.

Answer 42

Yayyyy cx

Answer 43

I guess I should advise, in case you didn't already, to fix part (a) if you did put b=-4 instead of b=40. That would be silly to mix up there. :)

Answer 44

there you go again helpin me x'D

Answer 45

so for the second one b would be -223?

Answer 46

third*

Answer 47

2x^2 + x + 28 = 0
a=2, b=1, and c=28
So 1^2 - 4*2*28 = -223. Looks good!

Answer 48

okie dokie i think i got it from here, ill check in when im completly done.

Answer 49

Ohh I finally figured out what I was seeing that looked so weird.
For some reason I am seeing the equation written twice on this document, side by side without spacing:
\(\color{red}{x^2 -4x - 5 = 0} \color{blue}{x^2 - 4x - 5 = 0} \)
\(x^2 - 4x - 5 = 0x^2 - 4x - 5 = 0 \) ???
That makes this seem a lot easier. lol.

Answer 50

yea i think because when the teachers send it, its sent sideways idk why though cx

Answer 51

how do we find out how many solutions it is again? Dx

Answer 52

Using the discriminant:
b^2 - 4ac > 0 ==> 2 real solutions
b^2 - 4ac = 0 ==> 1 real solution
b^2 - 4ac < 0 ==> 2 (imaginary) solutions

Answer 53

wow. that just made it 1000 times easier cx

Answer 54

1)
A|a=1 b=-4 c=-5
B|b^2 - 4ac = (-4)^2 - 4*(1)*(-5) = 36
C| two imaginary solutions.
2)
A| a=8 b=40 c=50
B| b^2-4ac=(-40)^2-4*(8)*(50)=0
C| one real solution
3)
A| a=2 b=1 c=28
B| b^2-4ac= (1)^2-4*(2)*(28)=-223
C|two real solutions.
4)
A| a=6 b=-2 c=5
B| b^2-4ac=(-2)^2-4*(6)*(5)=-116
C|two real solutions.
5)
A| a=1 b=8 c=-16
B| b^2-4ac=(8)^2-4*(1)*(-16)= -48
C|two real solutions.

hows this?

Answer 55

First thing I notice is you flipped around the condition for two real and two imaginary solutions. The discriminant (b^2 - 4ac) > 0 means two REAL solutions, and < 0 means two IMAGINARY solutions. For #1, you had discriminant of 36, but that means two REAL solutions because it is greater than 0.

Answer 56

so is everything else correct then?

Answer 57

The other thing is specifically for #5. Notice the equation is written:
\(x^2 = 8x - 16 \)
That is actually not in the form \(ax^2 + bx + c = 0\) yet! We need to get 0 on one side of the equation first, by moving everything else over with x^2.
We subtract 8x and add 16 to both sides.
\(x^2 \color{green}{-8x + 16} = 0 \)

Answer 58

Oh ok ok so ima put that and actually use that to solve ok cx

Answer 59

Yep. That is a really common mistake; they always try to pull that on students, mixing up the equation. You just need to check and make sure there is one side that is 0. That's not too much to inspect. :)

Answer 60

5) = x^2-8x+16=0
A| a=1 b=-8 c=16
B| b^2-4ac= (-8)^2-4*(1)*(16)= 0
C| one real solution

this is what i got

Answer 61

Looks good now!

Answer 62

thank you so much ^-^

Answer 63

Glad to help!

Answer 64

@AccessDenied are you on Dx