Question

How many real solutions are there to the equation shown below?

x2 + 2x + 6 = 0

Answer 1

to know, you'll have to factor ^_^

Answer 2

none:

the discriminate is negative:

4 -24 = -20

Answer 3

how come? .-.

Answer 4

for a quadratic equation ax^2 + bx+ c
Disc = b^2 -4ac
if D<0 roots are imaginary

Answer 5

that (b^2 -4ac) is negative

Answer 6

i qot this we i factored it .

Answer 7

right, the solutions are imaginary ;)

Answer 8

i was thinking it Was 2 real solutions .

Answer 9

if its got an "i" in it; that means "i"maginary. Its a cruel joke really, such a useful complex number and all, but to be dubbed imaginary all your life is kinda a misnomer

Answer 10

\[x=-1\pm \sqrt{-5}\]

Answer 11

it simply means it exists, but not in the realm of the "real" numbers. So it tends to not get invited to all the social gatherings and all the other numbers that hang out with it soon leave ...

Answer 12

oh, I missed that ^_^"

Answer 13

so i quess It foold Mee uqqh .

Answer 14

\[ax^2 + bx + c = 0 \]
\[\implies x=\frac{-b\pm \sqrt{b^2-4(a)(c)}}{2(a)}\]

\[b^2 - 4(a)(c) < 0 \implies \text{No real solutions}\]
Because the thing under the square root would be negative.
\[b^2 - 4(a)(c) = 0 \implies \text{One real solution}\]
Because the \(\pm\) bit would be \(\pm\) 0.
\[b^2 - 4(a)(c) > 0 \implies \text{Two real solutions}\]

Answer 15

in math, you have to use maths definitions...not english definitions lol

Answer 16

"real" in math means: not having an "i"

Answer 17

idk , I dont qet it Maybe one solution

Answer 18

\[x=-1\pm \sqrt{5 \times i ^{2}}=-1\pm i \sqrt{5}\]

Answer 19

No SOLUTIONS

Answer 20

no REAL solutions lol

Answer 21

If the part under the square root is negative you have no real soltions.

Answer 22

but how do yu find if theres Some solutions or not how yu determine

Answer 23

Look at the quadratic equation.

Answer 24

Have you learned the quadratic equation?

Answer 25

yea

Answer 26

Home sweet home:

Answer 27

Ok. So when you plug in a, b, and c into the quadratic equation. You see the part with the square root?

Answer 28

What is the domain of the function F(x) graphed below?

Answer 29

Ohh okayy i qet tht now .

Answer 30

the set of all Real numbers

Answer 31

The domain of the function is all the values you can give x and get real solutions for F(x). Since there are no vertical asymptotes or holes in the graph, the domain is \((-\infty,\infty)\)

Answer 32

amistre64 Wats tht yu wrote the set of all real numbers

Answer 33

-1 < x < 1

Answer 34

Yes; the domain of your graph there is the "set of all Real numbers".

All the numbers that live on the number line are in the domain

Answer 35

the domain for f(x) = (x-1)^2 -1 is; any and all real numbers. The graph extends forever to the left, and forever to the right with no restrictions

Answer 36

What is the y-coordinate of the vertex of a parabola with the following equation?

y = x2 - 8x + 18

Answer 37

(-b/2a, f(-b/2a) :)

Answer 38

x=4; y = 4^2 -8(4) + 18

Answer 39

16-32+18 = -16+18 = 2

(4,2)

Answer 40

itss 18 right

Answer 41

18 is the "y intercept" that is not the vertex for this graph.

Answer 42

the vertex for this graph is (4,2) so the y coordinate of the vertex is 2

Answer 43

the vertex is where the graph bends and starts going back around; its kinda like the "handle" of the graph. we can use it to move the graph anywhere we want of the graph paper

Answer 44

ohh Okay but I thought it was 18 because i factor it & i thouqht they was asking for tha y intercept But read it to fast & put tha wrong answer . dummie Mee

Answer 45

lol... I do that alot as well; I read what I want to see in it and get a wrong answer ;)