[Kimmitmelvirm]

I does Algebra, Trigonometry, Calculus, Differential Equations, and most Calculus-based Physics.

See how this goes, it helps me review old stuff too.

[Uttephabeta]

Can you Complete the Square?

In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form
to the form
In this context, "constant" means not depending on x. The expression inside the parenthesis is of the form (x − constant). Thus one converts ax2 + bx + c to
and one must find h and k.
Completing the square is used in

• solving quadratic equations,
• graphing quadratic functions,
• evaluating integrals in calculus,
• finding Laplace transforms.

In mathematics, completing the square is considered a basic algebraic operation, and is often applied without remark in any computation involving quadratic polynomials.

Of course I had to post that just incase you forgot what it was so you could refresh your memory.

[Agitoligflise]

I remember doing this, I can help

If you want, you can make a new thread with the problems that you need help with, and we can go from there.

[Fededorbprago]

I remember doing this, I can help

If you want, you can make a new thread with the problems that you need help with, and we can go from there.

I'll do that tomorrow when my teacher permits it

[chuecafressds]

I remember doing this, I can help

If you want, you can make a new thread with the problems that you need help with, and we can go from there.

No one was asking you for help sir. Get your own thread.

War Machine, if you'll post some problems up I'll try and help you out.

[Kilsimpaile]

i'll post some up tomorrow because my teacher is telling me to work on something else.

[oneliRafmeene]

Sorry for intruding, BP.

[TineSeign]

-deleted-

my bad, didnt read whole thread b4 posting.

[quack!]

Here's what I'm having trouble with:
Quadratic Formula;
-b2 +- square root (b2 – 4ac) / 2a



-5x2 – 3x – 3=0


I can't get the last part right :

A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real. The roots are given by the quadratic formula:
where the symbol "±" indicates that both
and are solutions.

In the above formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using an upper case Greek Delta:
A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case the discriminant determines the number and nature of the roots. There are three cases:

• If the discriminant is positive, there are two distinct roots, both of which are real numbers.

For quadratic equations with integer coefficients, if the discriminant is a perfect square, then the roots are rational numbers—in other cases they may be quadratic irrationals.

• If the discriminant is zero, there is exactly one distinct real root, sometimes called a double root:

• If the discriminant is negative, there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other:

and where i is the imaginary unit. Thus the roots are distinct if and only if the discriminant is non-zero, and the roots are real if and only if the discriminant is non-negative.

When it's actually this:

Quadratic factorization

The term
is a factor of the polynomial
if and only if r is a root of the quadratic equation
It follows from the quadratic formula that
In the special case (b2 = 4ac) where the quadratic has only one distinct root (i.e. the discriminant is zero), the quadratic polynomial can be factored as

Can you try and help me BP if you don't mind bro?

[NowFloabDow]

Given: -5x²-3x-3=0
Using the quadratic equation: -b±√(b²-4ac)/2a
→ 3±√(9-60)/-10
→ 3±√(-51)/-10

You cannot take the square root of a negative value, the only way to express it is using the imaginary value i, where i=√(-1). So you factor in the radical.

→ 3±√(-1*51)/-10
→ 3±√(-1)*√(51)/-10
→ 3±√(51)i/-10

Now you can evaluate two separate answers which will have i. Your answer should look like a+bi, where a and b are the real part and i is the imaginary part.

[Qncvqpgfg]

Given: -5x²-3x-3=0
Using the quadratic equation: -b±√(b²-4ac)/2a
→ 3±√(9-60)/-10
→ 3±√(-51)/-10

You cannot take the square root of a negative value, the only way to express it is using the imaginary value i, where i=√(-1). So you factor in the radical.

→ 3±√(-1*51)/-10
→ 3±√(-1)*√(51)/-10
→ 3±√(51)i/-10

Now you can evaluate two separate answers which will have i. Your answer should look like a+bi, where a and b are the real part and i is the imaginary part.

Thank you! You've helped me complete a few lessons on this subject on novanet bropaladin. I'm going to show my teacher how useful this place can be now during fourth block.

[Freefspruptpx]

This thread needs more assignments.

[RIjdrVs3]

I would give u some but you would never figure them out

[Xodvbooj]

I would give u some but you would never figure them out

just give him something so he won't get bored (plus this thread is why I have an entirely legitimate purpose for coming to this site during school hours).

[Toivaluadiora]

I would give u some but you would never figure them out

You're not smart enough to give me hard problems.

[Sironimoll]

lol u said hard

I've got some to throw at you that would take up most of your time

[paydayus]

Let's see them. I'm curious.

[Qahtwugc]

Given: -5x²-3x-3=0
Using the quadratic equation: -b±√(b²-4ac)/2a
→ 3±√(9-60)/-10
→ 3±√(-51)/-10

You cannot take the square root of a negative value, the only way to express it is using the imaginary value i, where i=√(-1). So you factor in the radical.

→ 3±√(-1*51)/-10
→ 3±√(-1)*√(51)/-10
→ 3±√(51)i/-10

Now you can evaluate two separate answers which will have i. Your answer should look like a+bi, where a and b are the real part and i is the imaginary part.

Hey Black Paladin, what if the denominator is positive, but the number in the square root is?

Here it is:
7x2 – 6x + 3 = 0

6±√(62 – 4 (7) (3)/2(7)

6±√(36-84)/2(7)

6±√(-48)/14

[Enfonebew]

If the number in the square root is negative it has the imaginary constant i

So just factor the -1 in the square root along with all the factors of 48.

=√(-48)
=√(48*-1)
=√(2*24*-1)
=√(2*2*12*-1)
=√(2*2*2*6*-1)
=√(2*2*2*2*3*-1)

You have four factors of 2, one factor of 3, and one factor of -1. When you have two factors of the same number under a radical you can pull that number out, since √(4)=√(2*2)=2. (It's not limited to numbers, if you have two variables that are the same and are being multiplied together you can do the same thing)

=2*2√(3*-1)
=4√(3)i

There's your imaginary part of the equation


Having a negative value in the radical is the what the concept of imaginary values are about. You represent the √(-1) as i by factoring the radical.

Also, having a positive or negative value in the denominator doesn't change anything about using the quadratic formula except sign changes.

[drugimpotence]

Hey just curious is thier a set forumla for working out the permatations in a sequence of numbers?