Solving the quadraric functions  TOPIC_SOLVED

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Solving the quadraric functions  TOPIC_SOLVED

Postby honest_denverco09 on Fri Apr 10, 2009 2:06 am

I would like to ask that there are how many ways (methods) we can use to solve the quadratic function (ax^2 + bx + c = 0) and which one is the fastest ?
Thank you!
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Re: Solving the quadraric functions

Postby Martingale on Fri Apr 10, 2009 2:11 am

honest_denverco09 wrote:I would like to ask that there are how many ways (methods) we can use to solve the quadratic function (ax^2 + bx + c = 0) and which one is the fastest ?
Thank you!


I just try and factor the quadratic...if i can't then I use the quadratic equation.
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Re: Solving the quadraric functions

Postby honest_denverco09 on Fri Apr 10, 2009 2:40 am

What do you mean with "factor" ? Did you mean that factor is the process to converse the general form to the factor form or you mean that factor is the process to change the general form to the square form ?

If you mean the second meaning, i think it is not necessary to do so, because when the quadratic function is in general form, it is more easier to solve.

To use the quadratic formula, it is the best way, but i want to know the other way, and i knew one way to do, but i don't know how.
It is the way that use the rectangular diagram to converse the general form into the factor form, then the middle terms (the square at the top right corner and the square at the bottom left corner) which are assigned by "a" and "b" in the previous topics, are our solution (x1, and x2).
Notice that our solutions have the opposite signs with the sign of "a" and "b". The problem is that i don't know how to find the "a" and "b" in some case. For instant : Solve this function : 4x^2 + 7x -1 = 0
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Re: Solving the quadraric functions

Postby bonnyann91 on Fri Apr 10, 2009 3:03 am

I like to use the quadratic formula the best which is x=-b(+ or -) the square root of b^2 - 4ac all over 2a. But, many people think that factoring is also really easy if it is possible. I also think that you could do completing the square. So, there are a couple of ways you could solve a problem like that. Hope it helps!
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Re: Solving the quadraric functions

Postby honest_denverco09 on Fri Apr 10, 2009 3:25 am

Ok. Now how can i solve this problem by using the Completing square (Using the rectangular diagram) : 4x^2 + 7x - 1 = 0
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Re: Solving the quadraric functions

Postby Martingale on Fri Apr 10, 2009 4:07 am

not sure why you would want to Complete the square since that is how you derive the quadratic equation. You might as well just use the Equation.
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Postby stapel_eliz on Fri Apr 10, 2009 12:15 pm

honest_denverco09 wrote:I would like to ask that there are how many ways (methods) we can use to solve the quadratic function (ax^2 + bx + c = 0) and which one is the fastest ?

The methods available are described here. The last part of the lesson gives some examples of how to determine, for a given quadratic, what would probably be the quickest way to go. :wink:
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Re: Solving the quadraric functions

Postby kaytlin93 on Fri Apr 10, 2009 3:30 pm

There are three possible methods in solving a quadratic function. The first method , which would probably be the simplest, is to factor. For example if you had x^2 + x - 6 the factored form would be (x + 3)(x - 2) and from there you would set each pair equal to zero and solve for x. The second method would be to use the quadratic formula which is x= -b (+or-) square root of b^2 - 4ac all over 2a. The third possible method would be to complete the square, you would take the coefficient in front of the b value and divide it by 2. I hope this helps :)
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Re: Solving the quadraric functions

Postby honest_denverco09 on Sat Apr 11, 2009 12:17 am

Thank you for your answers!
But maybe you have a small mistake in the third way.
kaytlin93 wrote:to complete the square, you would take the coefficient in front of the b value and divide it by 2.

If you take the coefficient in front of the "b" in ax^2 + bx + c = 0, what will you get ? Did you mean we will get the solution (x) ? In fact, we won't.

For "x^2 + x - 6 = 0", we have :

If we would like to complete the square to solve quadratic function in general form, we should square root the "c" (if c <0, we square root (-c)). In some case, the square root of "c" is not whole numbers, if so, we need to round it to the integer, for example (square root of 6 is 2.449... , we will round to 2). And also, the square root of "c" is one of our solution (x), we call x1st.

As above, assume that our solution (also our x-intercepts) are x1st and x2, we did have x1 = square root of "c" = 2. So how can we find x2nd. We already to know that (x1st)(x2nd) = c, so x2nd = c / x1st = -6 / 2 = -3.

Eventually, our solution are x = -3 or x = 2
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