Question

http://prntscr.com/l4zky7

Answer 1

@Vocaloid

Answer 2

You're trying to set it equal to 0, do you know how to do that?

Answer 3

im not sure

Answer 4

Add 1 on both sides

Answer 5

dont we have to find what a,b, and c is?

Answer 6

Yes but that is only true when it is equal to 0

Answer 7

so -2x+1 and -1+1?

Answer 8

Right, but you cant really add it to the left side so it remains as a constant
\(x^2-2x+1=0\)

Answer 9

Now you find the a,b and c

Answer 10

a=1
b=-2x
c=1

?

Answer 11

Close, there arent x values for the a,b,c

a = 1
b = -2
c = 1

Answer 12

so 1^2-(-2)+1

Answer 13

Are you trying to find the discriminant?

Answer 14

yes

Answer 15

Discriminant = \(b^2-4ac\)

Answer 16

-2^2-4*1*1

Answer 17

Yes

Answer 18

got it
how about this one
http://prntscr.com/l4ztnz

Answer 19

@dude

Answer 20

@Vocaloid

Answer 21

Well if we want two solutions hen thand discriminate b^2 - 4ac must be positive

So pick a large value for b and a small value for 4ac and check to see that the values you picked give a positive discrimination

Answer 22

so 5^2-4*2*3

Answer 23

it equals 1

Answer 24

and 1 is positive

Answer 25

is this right http://prntscr.com/l504oy
@Vocaloid

Answer 26

almost

the equation is b^2 - 4ac so b = 5, while a = 2 and c = 3 the way you've written it

it also asks for the quadratic equation so you'd plug in your a, b, and c into ax^2 + bx + c and just give them the final equation

Answer 27

got it
http://prntscr.com/l505rg

Answer 28

one real number solution means the discriminant must be 0

so try picking a, b, and c such that b^2 - 4ac = 0

Answer 29

would the first one be like this
2x^2+5x+3

Answer 30

yes

Answer 31

it would be http://prntscr.com/l50822

Answer 32

and for the second one can you give me an example? @Vocaloid

Answer 33

@Vocaloid