How many real solutions are there to the equation shown below?
x2 + 2x + 6 = 0

- anonymous

How many real solutions are there to the equation shown below?
x2 + 2x + 6 = 0

- chestercat

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- anonymous

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

- amistre64

none:
the discriminate is negative:
4 -24 = -20

- anonymous

how come? .-.

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## More answers

- anonymous

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

- amistre64

that (b^2 -4ac) is negative

- anonymous

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- anonymous

i qot this we i factored it .

- amistre64

right, the solutions are imaginary ;)

- anonymous

i was thinking it Was 2 real solutions .

- amistre64

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

- radar

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

- amistre64

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 ...

- anonymous

oh, I missed that ^_^"

- anonymous

so i quess It foold Mee uqqh .

- anonymous

\[ax^2 + bx + c = 0 \]
\[\implies x=\frac{-b\pm \sqrt{b^2-4(a)(c)}}{2(a)}\]
\[b^2 - 4(a)(c) < 0 \implies \text{No real solutions}\]
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.
\[b^2 - 4(a)(c) > 0 \implies \text{Two real solutions}\]

- amistre64

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

- amistre64

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

- anonymous

idk , I dont qet it Maybe one solution

- radar

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

- amistre64

No SOLUTIONS

- amistre64

no REAL solutions lol

- anonymous

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

- anonymous

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

- anonymous

Look at the quadratic equation.

- anonymous

Have you learned the quadratic equation?

- anonymous

yea

- amistre64

Home sweet home:

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- anonymous

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- anonymous

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

- anonymous

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

- anonymous

Ohh okayy i qet tht now .

- amistre64

the set of all Real numbers

- anonymous

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 \((-\infty,\infty)\)

- anonymous

amistre64 Wats tht yu wrote the set of all real numbers

- anonymous

-1 < x < 1

- amistre64

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

- amistre64

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

- anonymous

What is the y-coordinate of the vertex of a parabola with the following equation?
y = x2 - 8x + 18

- amistre64

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

- amistre64

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

- amistre64

16-32+18 = -16+18 = 2
(4,2)

- anonymous

itss 18 right

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

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