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Explain the methods for solving quadratics. Do you understand the method / concept of completing the square and why we use it? Describes the roots of a quadratic equation. in what type of work a are quadratic equation used?

Thank you for the opportunity to help you with your question!

More help this is the link.........................http://www.purplemath.com/modules/quadform.htm

The quadratic in
the previous section's last
example,
"(*x* – 2)^{2}
– 12", can be multiplied
out and simplified to be "*x*^{2} – 4*x* – 8".
But we would not have been able to solve the equation with the quadratic
formatted this way because it doesn't factor and it isn't ready for square-rooting.
The only reason we could solve it before was because they'd already put
all the *x* stuff inside a square, so we could square-root both sides. So how do you
go from a regular quadratic like "*x*^{2} – 4*x* – 8" to one that is
ready to be square-rooted? We would have to "complete the square".

I have a lesson on solving quadratics by completing the square, which explains the steps and gives examples of this process. It also shows how the Quadratic Formula is generated by this process. So I'll just do just one example of the process in this lesson. If you need further instruction, read the lesson at the above hyperlink.

**Use****completing the square to solve***x*^{2}– 4*x*– 8 = 0.

**Solve***x*^{2}+ 3*x*– 4 = 0**Solve 2***x*^{2}– 4*x*– 3 = 0. Round your answer to two decimal places, if necessary.

As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the quadratic in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that form, and then solve. It works like this:

First, I put the loose number on the other side of the equation:

*x*^{2} – 4*x* – 8 = 0

*x*^{2} – 4*x* = 8
Copyright
© Elizabeth Stapel 2002-2011 All Rights Reserved

Then I look at
the coefficient of the *x*-term,
which is –4 in this case. I take half of this number (including the sign), giving
me –2.
Then I square this value to get +4,
and add this squared value to both sides of the equation:

*x*^{2} – 4*x* + 4 = 8 + 4

*x*^{2} – 4*x* + 4 = 12

This process creates a quadratic that is a perfect square, and factoring gives me:

(*x* – 2)^{2} = 12

Now I can square-root both sides of the equation, simplify, and solve:

(*x* – 2)^{2} = 12

**Then
the solution is **

Unless you're told that
you *have* to use completing the square, you will probably not use
this method, in actual practice, when solving quadratic equations. Either
some other method (such as factoring) will be obvious and quicker, or
else the Quadratic
Formula (coming
up next)
will be easier to use. However, if your class covered completing the square,
you should expect to be required to show that you can complete the square
to solve a quadratic on the next test.

Often, the simplest way
to solve "*ax*^{2} + *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 "*ax*^{2} + *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 |

For the Quadratic Formula
to work, you *must* have your equation arranged in the form "(quadratic)
= 0".
Also, the "2*a*"
in the denominator of the Formula is underneath *everything* above,
not just the square root. And it's a "2*a*"
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 "*b*^{2}"
means "the square of ALL of *b*,
including its sign", so don't leave *b*^{2} 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!

Here are some examples of how the Quadratic Formula works:

This quadratic happens to factor:

*x*^{2} + 3*x* – 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:

Then, as expected, the
solution is ** x = –4, x = 1**.

Suppose you have *ax*^{2} + *bx* + *c* = *y*,
and you are told to plug zero in for *y*.
The corresponding *x*-values
are the x-intercepts of the graph. So solving *ax*^{2} + *bx* + *c* = 0 for *x* means, among other things, that you are trying to find *x*-intercepts.
Since there were two solutions for *x*^{2} + 3*x* – 4 = 0,
there must then be two *x*-intercepts
on the graph. Graphing, we get the curve below:

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As you can see, the *x*-intercepts
(the red dots above) match the solutions, crossing the *x*-axis
at

*x* = –4 and *x* = 1. This shows the
connection between graphing and solving: When you are solving "(quadratic)
= 0", you are finding the *x*-intercepts
of the graph. This can be useful if you have a graphing calculator, because
you can use the Quadratic Formula (when necessary) to solve a quadratic,
and then use your graphing calculator to make sure that the displayed *x*-intercepts
have the same decimal values as do the solutions provided by the Quadratic
Formula.

Note, however, that the calculator's display of the graph will probably have some pixel-related round-off error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

There are no factors
of (2)(–3)
= –6 that add up
to –4,
so I know that this quadratic cannot be factored.
I will apply the Quadratic Formula. In this case, *a* = 2, *b* = –4, and *c* = –3:

Then the answer is *x* = –0.58, *x* = 2.58,
rounded to two decimal places.

Warning: The "solution"
or "roots" or "zeroes" of a quadratic are usually
required to be in the "exact" form of the answer. In the example
above, the exact form is the one with the square roots of ten in it. You'll
need to get a calculator approximation in order to graph the *x*-intercepts
or to simplify the final answer in a word problem. But unless you have
a good reason to think that the answer is *supposed* to be a rounded answer, always go with the exact form.

Please let me know if you need any clarification. I'm always happy to answer your questions.

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