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Dec 6, 2013
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Dec 6, 2013
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Dec 6, 2013
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Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. While factoring may not always be successful, the Quadratic Formula can always find the solution.

The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve. The Quadratic Formula is derived from the process of completing the square, and is formally stated as:

For ax2 + bx + c = 0, the value of x is given by:





x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.

In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this! These are just some examples of using the formula!

Here are some examples of how the Quadratic Formula works:

Solve x2 + 3x – 4 = 0

This quadratic happens to factor:

x2 + 3x – 4 = (x + 4)(x – 1) = 0

...so I already know that the solutions are x = –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:

x = -4, x = 1

Then, as expected, the solution is x = –4, x = 1.
Dec 6, 2013
 #3
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Hello all,
It occurs to me that perhaps I should also introduce myself.
I have a Bachelor of Science (in mathematics) and a Diploma of Education in secondary mathematics. My mathematics results were very good. My teacher training results were excellent.
I have spent many years teaching in seconday schools and I also taught for a number of years in a University College.
I have also tutored a great many students on an individual basis and have had consistently excellent results.

I 'discovered' this forum on 28th September and became a member immediately.
At that time almost all all post were going unanswered and the level of the questions tended to be very low.
I took it upon myself to see that all mathematics questions were answered. (there would be very few exceptions anyway)
Now, the quality of the questions has improved markedly and the forum is just starting to attract some really enthusiastic budding mathematicians.
Now I actually try not to answer all the questions. Although I probably answer more than I should. I believe that it is very important that the forum users are given the opportunity and encouragement to offer solutions and other input.
I have a passion for mathematics and an even greater passion for nuturing mathematical learning and mathematical appreciation in others.
I hope the other forum users believe that I have had a positive impact. If anyone has had questions I don't think that they have been afraid to ask. That is what I, and others, are here for.

In the coming months I hope the forum users will start to develop a more interactive approach. This has been slow in coming but it is just starting to happen now.
Melody.
Dec 6, 2013
 #4
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This is a very difficult question Rosey, I am wondering why your brother didn't ask it for himself.
I am thinking that he is just curious and that it isn't really a school question.
You have to use calculus to answer this question.
I will run through the working for you and your brother.
First i will assume that the car is travelling horizontally when it comes off the top of the cliff.
So it's vertical velocity (speed) is 0
It's horizontal velocity is 10 miles per hour = 44/3 feet/second

Now there are 5280 feet in a mile
Gravity will make the car accelerate as it falls
Gravity = 32.176 feet/second 2

So what you have to do is work out how long it will take the car to fall to the water.
After doing lots of calculations I got 1.5768 seconds ( approximately)
Now, how far will the car travel horizontally in this amoount of time.
44/3 feet/second * 1.5768 = 23 feet approximately.

If your brother has done some calculus and wants more help with the mathematics then he or you can post again and I will try to help more.
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Translation.
Espero que esta traducción de Bing tiene sentido.

Esta es una pregunta muy difícil Rosey, me pregunto por qué tu hermano no pedirlo mismo.
Estoy pensando que es sólo por curiosidad y que no es realmente una cuestión de escuela.
Tienes que usar cálculo para responder a esta pregunta.
Correré a través del trabajo para ti y para tu hermano.
Primero voy a suponer que el auto está viajando horizontalmente cuando se trata de la parte superior del acantilado.
Así es vertical velocity (velocidad) es 0
Su velocidad horizontal es 10 millas por hora = 44/3 pies/segundo

Ahora hay 5280 pies en una milla
Gravedad hará que el coche acelera como cataratas
Gravedad = 32,176 pies/segundo [sup] 2 [primera]

Así que lo que tienes que hacer es resolver Cuánto tiempo tardará el coche al caer al agua.
Después de hacer un montón de cálculos tengo 1,5768 segundos (aproximadamente)
Ahora, qué tan lejos el coche viajará horizontalmente en este amoount del tiempo.
44/3 pies / segundo * 1.5768 = 23 metros aproximadamente.

Si tu hermano ha hecho algunos cálculos y quiere más ayuda con las matemáticas entonces él o publique otra vez y voy a intentar ayudar más.
Dec 6, 2013

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