Algebra 2 Help pleaseee

Quote:

What are the solutions of 3x2 + 6x + 6 = 0?

just plug and chug using the quadratic formula

ax ² + bx + c = 0 ==> x = (1/2a)(-b +/- sqrt(b ²-4ac))

In this case a=3, b=6, c=6

plugging and chugging we get

x = (1/2*3)(-6 +/- sqrt(6 ²-4*3*6))
x = (1/6)(-6 +/- sqrt(36-72))
x = (1/6)(-6 +/- sqrt(-36))
x = (1/6)(-6 +/- 6i)
x = -1 +/- i

Could have divided by 3 first to make it a little earlier.

Thank you you to both of you, it turns out I had it in me because I got it right(: Thank you though.

But now I need help with a review question:

g(x) = 2x^2 −16x + 15 I need to know how to put that into standard form of a parabola which is g(x)= a(x − h)^2 + k if you can help(:

Quote:

g(x) = 2x^2 −16x + 15 I need to know how to put that into standard form of a parabola which is g(x)= a(x − h)^2 + k if you can help(:

this is called completing the square.
learn it, love it

2x ² - 16x + 15 ; factor the x ² coefficient to the outside

2[size=150]([/size]x ² - 8x + 15/2[size=150])[/size] ; now note the coefficient of the x term is 8 and we take 1/2 of this to be h in our (x-h) ² term

2[size=150]([/size](x-4) ² - 4 ² + 15/2[size=150])[/size] ; study this a sec we added and subtracted 4 ², make sure you understand how

2[size=150]([/size](x-4) ² - 16 + 15/2[size=150])[/size] ; compute the square of 4

2[size=150]([/size](x-4) ² - 17/2[size=150])[/size] ; do the addition

2(x-4) ² - 17 ; distribute the multiplcation by 2

Rom:

Quote:

g(x) = 2x^2 −16x + 15 I need to know how to put that into standard form of a parabola which is g(x)= a(x − h)^2 + k if you can help(:

this is called completing the square.
learn it, love it

2x ² - 16x + 15 ; factor the x ² coefficient to the outside

2[size=150]([/size]x ² - 8x + 15/2[size=150])[/size] ; now note the coefficient of the x term is 8 and we take 1/2 of this to be h in our (x-h) ² term

2[size=150]([/size](x-4) ² - 4 ² + 15/2[size=150])[/size] ; study this a sec we added and subtracted 4 ², make sure you understand how

2[size=150]([/size](x-4) ² - 16 + 15/2[size=150])[/size] ; compute the square of 4

2[size=150]([/size](x-4) ² - 17/2[size=150])[/size] ; do the addition

2(x-4) ² - 17 ; distribute the multiplcation by 2

Thank you so much!!! It helped me find the vertex for this function, with both way and now I know how to do it(:

My next and last question is this one:
Chelsea is trying to find the equation of a quadratic that has a focus of (1, 3) and a directrix of y = −3. Describe to Chelsea your preferred method for deriving the equation. Make sure you use Chelsea's situation as a model to help her understand.

You dont have to actually answer it cause I want to learn to do it myself, but could you give me some pointers on how to figure it out? I don't even understand what a directrix or a focus is/: I know it has to do with parabolas, but that's about it.

Can someone please help me? I know Ive asked a lot of questions here lately lol but pleasee

My next and last question is this one:
Chelsea is trying to find the equation of a quadratic that has a focus of (1, 3) and a directrix of y = −3. Describe to Chelsea your preferred method for deriving the equation. Make sure you use Chelsea's situation as a model to help her understand.

You dont have to actually answer it cause I want to learn to do it myself, but could you give me some pointers on how to figure it out? I don't even understand what a directrix or a focus is/: I know it has to do with parabolas, but that's about it.

And there is also this one:
Sandra exclaims that her quadratic with a discriminant of −4 has no real solutions. Sandra then puts down her pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Sandra, in calm and complete sentences, how to find the solutions, even though they are not real.

I know when the discriminant is negative it has two rational solutions, but it's telling me hers doesn't so it's confusing me a little???

Warped Tour Girl:

Can someone please help me? I know Ive asked a lot of questions here lately lol but pleasee

My next and last question is this one:
Chelsea is trying to find the equation of a quadratic that has a focus of (1, 3) and a directrix of y = −3. Describe to Chelsea your preferred method for deriving the equation. Make sure you use Chelsea's situation as a model to help her understand.

You dont have to actually answer it cause I want to learn to do it myself, but could you give me some pointers on how to figure it out? I don't even understand what a directrix or a focus is/: I know it has to do with parabolas, but that's about it.

Hi warped Tour Girl,
Consider a circle with a radius of 3 units and a centre (0,0)
You could say that this is the set of all points (x,y) such that they are all exactly 3 units from the point (0,0)
Using the distance formula, distance (0,0) to (x,y) is
sqrt [ (x-0)² + (y-0)² ]
= sqrt [ x² + y² ]
but we have already been told in the question that this is equal to 3 that is
sqrt [ x² + y² ] = 3
squaring both sides we get
x² + y² = 9 You have just found the equation of this circle from first principles.
This circle is the LOCUS of all points 3 units from (0,0)

Now, by definition, a parabola is the locus of all points equidistant from a line (the directrix) and a point (the focus)
your parabola has a focus of (1, 3) and a directrix of y = −3
Let (x,y) be a point on this function.
So
Perpendicular dist from (x,y) to y=-3 is equal to the distance of (x,y) to (1,3)
| y--3| = sqrt[ (x-1)²+(y-3)² ]
Squaring both sides we have,
(y+3)² = (x-1)²+(y-3)²
Now you need to expand and simplify and you will have your parabola.

I just found this neat little web page. Showing different parabola equations. You choose the directrix and the focus and it will draw the parabola and it will give you the equation.
Parabolas drawn from the directrix and the focus are best written in the form
(x-h)² = 4a(y-k)² where (h,k) is the vertex and a is the focal length. if a is pos then it is concave up and if it is neg then it is concave down.
I am probably getting a bit carried away here. Have a look at the web page.
https://www.khanacademy.org/math/trigonometry/conics_precalc/parabolas_precalc/e/parabola_intuition_2



And there is also this one:
Sandra exclaims that her quadratic with a discriminant of −4 has no real solutions. Sandra then puts down her pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Sandra, in calm and complete sentences, how to find the solutions, even though they are not real.

I know when the discriminant is negative it has two rational solutions, but it's telling me hers doesn't so it's confusing me a little???

I think you are confused a lot

NO the discriminant is the bit underneath the square root.
If it is negative the answer would have to be imaginary, you cannot find the square root of a negative number (not in the real number world anyway)
If the discriminant is negative there will be no real roots !!!
Can you have a go at it now?
Melody.

[/b][/color]
And there is also this one:
Sandra exclaims that her quadratic with a discriminant of −4 has no real solutions. Sandra then puts down her pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Sandra, in calm and complete sentences, how to find the solutions, even though they are not real.

I know when the discriminant is negative it has two rational solutions, but it's telling me hers doesn't so it's confusing me a little???[/quote] I think you are confused a lot

NO the discriminant is the bit underneath the square root.
If it is negative the answer would have to be imaginary, you cannot find the square root of a negative number (not in the real number world anyway)
If the discriminant is negative there will be no real roots !!!
Can you have a go at it now?
Melody.[/quote]

What I got to this question really confused me/: this is what I came up with for an answer:
b to the 2nd minus 4ac is the equation for the discriminant.
the function I've come up with is x^2-6x+13. Now we can plug it in:
6 to the 2nd is 36, -4 x (1)(13) = -48
so 36-64 is -16, even though the discriminant is negative, it doesn't mean it doesn't have real solutions, the discriminant when negative means it has 2 rational solutions, or real solutions.
To find the solutions start by isolating the terms with x in them.
So x^2-6x-13 subtract
x^2-6x=-13

Divide the coefficient on x, which is -6, by 2 and square the result:
and add to each side:
The answer you come up with should be 9, -6 divided by 2 is -3, -3 squared is 9.
Your equation should now look like this:
x^2-6x+9=-13+9 =-4
Now the perfect square trinomial on the left can be factored into a squared binomial.
Square x^2 and 9:
(x-3)^2=-4

Now solve:
square root both sides
x-3=-+ _/-4
x-3=+2
x=5
x-3=-2
x=1 Two real solutions!

Check your solutions:
(5)^2-6(5)+13=2 25-35+13=2 38-35=2 3=2 Irrational solution...
(1)^2-6(1)+13=-2 1-6+13=-2 7=-2 also incorrect..

So I'm still really confused

Warped Tour Girl:

[/b][/color]
And there is also this one:
Sandra exclaims that her quadratic with a discriminant of −4 has no real solutions. Sandra then puts down her pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Sandra, in calm and complete sentences, how to find the solutions, even though they are not real.

I know when the discriminant is negative it has two rational solutions, (NO! it has no rational roots) but it's telling me hers doesn't so it's confusing me a little???

I think you are confused a lot

NO the discriminant is the bit underneath the square root.
If it is negative the answer would have to be imaginary, you cannot find the square root of a negative number (not in the real number world anyway)
If the discriminant is negative there will be no real roots !!!
Can you have a go at it now?
Melody.[/quote]

What I got to this question really confused me/: this is what I came up with for an answer:
b to the 2nd minus 4ac is the equation for the discriminant.
the function I've come up with is F(x)=x^2-6x+13. Now we can plug it in:
6 to the 2nd is 36, -4 x (1)(13) = -48 (-52 actually)
so 36-64 (36-52 actually) is -16, even though the discriminant is negative, it doesn't mean it doesn't have real solutions, (yes it does) the discriminant when negative means it has 2 rational solutions (no it doesn't, you cannot find the square root of a negative number in the real realm), or real solutions.
To find the solutions start by isolating the terms with x in them.
So x^2-6x-13 subtract
x^2-6x=-13

Divide the coefficient on x, which is -6, by 2 and square the result:
and add to each side:
The answer you come up with should be 9, -6 divided by 2 is -3, -3 squared is 9.
Your equation should now look like this:
x^2-6x+9=-13+9 =-4
Now the perfect square trinomial on the left can be factored into a squared binomial.
Square x^2 and 9:
(x-3)^2=-4

Now solve:
square root both sides
x-3=-+ _/-4
x-3=+2
x=5
x-3=-2
x=1 Two real solutions!

Check your solutions:
(5)^2-6(5)+13=2 25-35+13=2 38-35=2 3=2 Irrational solution...
(1)^2-6(1)+13=-2 1-6+13=-2 7=-2 also incorrect..

So I'm still really confused[/quote] Yes I can see that

Try to come to terms with the stuff that i have put in red. After that you can attempt the real task.