anonymous
  • anonymous
How many real solutions are there to the equation shown below? x2 + 2x + 6 = 0
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
to know, you'll have to factor ^_^
amistre64
  • amistre64
none: the discriminate is negative: 4 -24 = -20
anonymous
  • anonymous
how come? .-.

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anonymous
  • anonymous
for a quadratic equation ax^2 + bx+ c Disc = b^2 -4ac if D<0 roots are imaginary
amistre64
  • amistre64
that (b^2 -4ac) is negative
anonymous
  • anonymous
anonymous
  • anonymous
i qot this we i factored it .
amistre64
  • amistre64
right, the solutions are imaginary ;)
anonymous
  • anonymous
i was thinking it Was 2 real solutions .
amistre64
  • 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
  • radar
\[x=-1\pm \sqrt{-5}\]
amistre64
  • 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
  • anonymous
oh, I missed that ^_^"
anonymous
  • anonymous
so i quess It foold Mee uqqh .
anonymous
  • 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
  • amistre64
in math, you have to use maths definitions...not english definitions lol
amistre64
  • amistre64
"real" in math means: not having an "i"
anonymous
  • anonymous
idk , I dont qet it Maybe one solution
radar
  • radar
\[x=-1\pm \sqrt{5 \times i ^{2}}=-1\pm i \sqrt{5}\]
amistre64
  • amistre64
No SOLUTIONS
amistre64
  • amistre64
no REAL solutions lol
anonymous
  • anonymous
If the part under the square root is negative you have no real soltions.
anonymous
  • anonymous
but how do yu find if theres Some solutions or not how yu determine
anonymous
  • anonymous
Look at the quadratic equation.
anonymous
  • anonymous
Have you learned the quadratic equation?
anonymous
  • anonymous
yea
amistre64
  • amistre64
anonymous
  • anonymous
anonymous
  • anonymous
Ok. So when you plug in a, b, and c into the quadratic equation. You see the part with the square root?
anonymous
  • anonymous
What is the domain of the function F(x) graphed below?
anonymous
  • anonymous
Ohh okayy i qet tht now .
amistre64
  • amistre64
the set of all Real numbers
anonymous
  • 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
  • anonymous
amistre64 Wats tht yu wrote the set of all real numbers
anonymous
  • anonymous
-1 < x < 1
amistre64
  • 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
  • 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
  • anonymous
What is the y-coordinate of the vertex of a parabola with the following equation? y = x2 - 8x + 18
amistre64
  • amistre64
(-b/2a, f(-b/2a) :)
amistre64
  • amistre64
x=4; y = 4^2 -8(4) + 18
amistre64
  • amistre64
16-32+18 = -16+18 = 2 (4,2)
anonymous
  • anonymous
itss 18 right
amistre64
  • amistre64
18 is the "y intercept" that is not the vertex for this graph.
amistre64
  • amistre64
the vertex for this graph is (4,2) so the y coordinate of the vertex is 2
amistre64
  • 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
  • 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
  • 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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