Question

solve each of the quadratic equations below and describe what the solutions represent to the graph of each: y=x^2 + 3x + 2
y=x^2 +2x + 1

Answer 1

help

Answer 2

Factor each function and check for which x values it equals 0.

Answer 3

for the first equation i am getting x= -1/4 Can that be correct?

Answer 4

Are these two separate problems or is this a system of quadratic equations?

Answer 5

Separate problems @mathstudent55

Answer 6

two separate problems to describe what solutions represent to the graph of each. I got X= -1/4 and I am not sure that can be correct.

Answer 7

\[y=x^2+3x+2 \rightarrow y=(x+1)(x+2) \rightarrow (x+1)(x+2)=0 \implies x = -1,-2\]Get what I did here? Now try doing the second one.

@mathgirl2012

Answer 8

Ok, set each right side equal to zero.
y=x^2 + 3x + 2
So do x^2 + 3x + 2 = 0
Factor the left side, and solve for x.

Answer 9

Having any success? @mathgirl2012

Answer 10

I did that and got x=1 x=2 then I solved for the vertex x=-b/2a and I got x+-1/4?

Answer 11

Then do the same for the second equation:
y=x^2 +2x + 1
Set x^2 + 2x + 1 = 0
Factor the left side, and solve for x.

Answer 12

For the first one, the solutions are x = -2 or x = -1

Answer 13

second equation i got x=1

Answer 14

For first eq, -b/2a = -3/2 = -1.5

Answer 15

The vertex is at x = -1.5

Answer 16

After factoring, be careful.
The firsrt equation factors into
(x + 2)(x + 1) = 0
Next step is:
x + 2 = 0 or x + 1 = 0, so you get
x = -2 or x = -1
The answers are NEGATIVE 2 and NEGATIVE 1, not 2 and 1

Answer 17

Second equation is:\[y=x^2+2x+1 \rightarrow y = (x+1)^2 \rightarrow (x+1)^2=0 \implies x = -1\]

Answer 18

Read my explanation just above.

Answer 19

For the first equation I got x=1, x=2 how did you get x=-2 , x= -1 when the equation reads
x^2 + 3x + 2??? @mathstudent55

Answer 20

Thank you

Answer 21

@mathgirl2012 I just answered. Read my post just a little above.

Answer 22

Because when you factor it, you get y = (x + 1)(x + 2). Then when you are finding solutions, you set the it to equal 0 like this: (x + 1)(x + 2) = 0. Now you find x-values where the function is 0. You know that if the first bracket is 0, then the whole thing is 0, if the second bracket 0, then the whole function is also.

So if x + 1 = 0, then x = -1. If x + 2 = 0, then x = -2. Therefore, the zeroes are x = -1, -2 because at these values, the whole function becomes 0.

@mathgirl2012

Answer 23

x^2 + 3x + 2 = 0
(x + 2)(x + 1) = 0
x + 2 = 0 or x + 1 = 0
x = -2 or x = -1

Answer 24

Now for the vertex, which has x = -b/2a, use the a, b, and c of the original equation.
You have x^2 + 3x + 2 = 0
When comparred with ax^2 + bx + c = 0, you see that a = 1, b = 3, and c = 2, so
-a/2b = -2/3 = -1.5
The vertex has an x-coordinated of -1.5

Answer 25

your the best you helped me and i wish i could give you both medals @mathstudent55 and @genius12

Answer 26

Awwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww. You can Fan me =D lol

Answer 27

Thanks. I'll take care of genius12

Answer 28

lol <3 @mathstudent55

Answer 29

@genius12 and @mathstudent55 you guys can fan me 2 lol ;P