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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to know, you'll have to factor ^_^

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0none: the discriminate is negative: 4 24 = 20

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for a quadratic equation ax^2 + bx+ c Disc = b^2 4ac if D<0 roots are imaginary

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that (b^2 4ac) is negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i qot this we i factored it .

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0right, the solutions are imaginary ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i was thinking it Was 2 real solutions .

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh, I missed that ^_^"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i quess It foold Mee uqqh .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ax^2 + bx + c = 0 \] \[\implies x=\frac{b\pm \sqrt{b^24(a)(c)}}{2(a)}\] \[b^2  4(a)(c) < 0 \implies \text{No real solutions}\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0in math, you have to use maths definitions...not english definitions lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0"real" in math means: not having an "i"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0idk , I dont qet it Maybe one solution

radar
 5 years ago
Best ResponseYou've already chosen the best response.0\[x=1\pm \sqrt{5 \times i ^{2}}=1\pm i \sqrt{5}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no REAL solutions lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the part under the square root is negative you have no real soltions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but how do yu find if theres Some solutions or not how yu determine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Look at the quadratic equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you learned the quadratic equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok. So when you plug in a, b, and c into the quadratic equation. You see the part with the square root?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the domain of the function F(x) graphed below?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohh okayy i qet tht now .

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the set of all Real numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0amistre64 Wats tht yu wrote the set of all real numbers

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Yes; 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
 5 years ago
Best ResponseYou've already chosen the best response.0the domain for f(x) = (x1)^2 1 is; any and all real numbers. The graph extends forever to the left, and forever to the right with no restrictions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the ycoordinate of the vertex of a parabola with the following equation? y = x2  8x + 18

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=4; y = 4^2 8(4) + 18

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01632+18 = 16+18 = 2 (4,2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.018 is the "y intercept" that is not the vertex for this graph.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the vertex for this graph is (4,2) so the y coordinate of the vertex is 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0ohh 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
 5 years ago
Best ResponseYou've already chosen the best response.0lol... I do that alot as well; I read what I want to see in it and get a wrong answer ;)
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