anonymous
  • anonymous
What is the quadratic formula? What is it used for? Give an uncommon example.
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
this one?
anonymous
  • anonymous
It's one of three my friend! lol.
amistre64
  • amistre64
an uncommon example eh..... the quadratic formula is derived from a method know as "completing the square"; and is used to find the real roots of a quadratic expression (quadratic just being a fancy name for "squared").

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amistre64
  • amistre64
there are 2 named parts to the quadratic formula that I tend to get backwards; the determinate and the discriminate..they were haveing a sale on "d" words back then and so were stuck with them
anonymous
  • anonymous
lol....ok
amistre64
  • amistre64
the discriminte, if it so be called, is the part -b/2a and it gives you the axis of symmetry of the parabola.... to find the vertex, the hoghest or lowest point of the parabola; we use the vertex in the equation to come up with the "y" part...make sense so far?
amistre64
  • amistre64
the other "d" word, the determinate perhaps, is a square root portion of the quad formula and it tells us whether there is: no roots, 1 root, or 2 roots.
anonymous
  • anonymous
ummmm........clear as mud. But it's mainly me.
anonymous
  • anonymous
So that would be the uncommon example?
amistre64
  • amistre64
dunno about any "uncommon" examples really, still trying to figure out what that means... maybe its talking about a quadratic that aint easily factored.....
amistre64
  • amistre64
if the square root portion has under its radical umbrella a negative #, there are no real roots; if it equals zero, then there is only 1 root, and if it is greater than zero, there are going to be 2 roots.
amistre64
  • amistre64
the quad formula is: -b sqrt(b^2 - 4ac) ---- +- ------------- 2a 2a here you can see the two parts :) i shoulda prolly put that up first...
anonymous
  • anonymous
No just an everyday example that might not be the first one you think of. Like, doesn't airplanes have something to do with the quadratic formula?
anonymous
  • anonymous
SO that problem has two roots?
amistre64
  • amistre64
that is not a problem ; that is the formula to determine the roots :) the standard quadratic equation is of the form: ax^2 +bx +c where a.b.and c are numbers...coefficients..
anonymous
  • anonymous
Lol....wow.
amistre64
  • amistre64
depending on the values of a,b,and c we can use the quadratic formula to determine its axis of symmetry, vertex coordinates, and all known solutions to y=0
amistre64
  • amistre64
lets try to make up an equation and test the quad formula out.... and try to stay away from airplanes :)
amistre64
  • amistre64
what are your favorite three numbers in the whole wide world... :)
anonymous
  • anonymous
3?
amistre64
  • amistre64
3 numbers.....yeah
anonymous
  • anonymous
Oh wait.......it said provide a USEFUL example..... duh. Not uncommon, or whatever I wrote.
anonymous
  • anonymous
7, 3, 35
amistre64
  • amistre64
ok..7 3 and 35; lets put them into the hat....stir it up...and pick one...............3 ok, a = 3; try again................7; b = 7 .......and stir again for effect........we get a 35; c = 35
amistre64
  • amistre64
3x^2 +7x -35...thats a quadratic formula...is it useful? dunno. but lets try out our formula.
amistre64
  • amistre64
the first thing we should do is see if its worth trying to find any roots: sqrt((7)^2 - (4)(3)(-35)) if this is zero or above, were good to go. sqrt(49 + 420) = sqrt(469) its a keeper so that last part is gonna be: +- sqrt(469)/2(3) the axis of symmetry is gonna be: -b/2a = -7/2(3) = -7/6
amistre64
  • amistre64
so we have a center for our parabola, its at x = -7/6 if we add and subtract the other part from this, we will know our solutions.
anonymous
  • anonymous
Dear God my brain hurts.
anonymous
  • anonymous
Any useful examples to share?
amistre64
  • amistre64
(-7/6) + sqrt(469)/6 = about 2.4427
amistre64
  • amistre64
this might be useful :) one example is the arch formed by throwing a baseball to your son.... you can determine its position with a quadratic equation.... but the "roots" of it are not that exciting :)
anonymous
  • anonymous
An arch of throwing a baseball?
amistre64
  • amistre64
the other root we got is about: -4.7761 you never seen a person throw a baseball around? it tends to go up, then comes back down into the golve of the oter person. makes for a nice arch while it flys thru the air
anonymous
  • anonymous
Yeah! Gotcha! Now......what do you know about Pythogorean Therorem's......or whatever it is?
amistre64
  • amistre64
i know that guy was a nutjob..... but his theorum is pretty helpful :)
anonymous
  • anonymous
So was Pavlov. Ok......one sec...
anonymous
  • anonymous
There are many applications for the Pythagorean Theorem.. can you give a specific example and show with calculations how the Pythagoras theorem applies?
amistre64
  • amistre64
this is the pyth thrm....
1 Attachment
amistre64
  • amistre64
the square of the sum of the legs of a right trianlge are equal to the square of the hypotenuse side.
amistre64
  • amistre64
a^2 + b^2 = c^2
anonymous
  • anonymous
OH.....and one more on quadratics too..
anonymous
  • anonymous
Nice attachment, am I supposed to color it? JK!
amistre64
  • amistre64
lol..... use it as a place mat ;)
anonymous
  • anonymous
So that a^2 thing is the formula for that>?
amistre64
  • amistre64
it is.... the names of the variables are unimportant, but that is a common rendition of it. another one is: x^2 + y^2 = r^2 and another one is: sin^2 + cos^2 = 1
anonymous
  • anonymous
all for the sum of the legs of a right triangle?
amistre64
  • amistre64
it can even be expanded to: c^2 = a^2 + b^2 -2ab cos(C)
amistre64
  • amistre64
yep, since right triangle are about the easiest to use; its nice to have stuff you can use with them :)
amistre64
  • amistre64
since the cos(90) = 0, that last part vanishes in the right trianlge and your left with the sum of the squares of the sides :)
anonymous
  • anonymous
Ok......one more the quadratics thing.....
amistre64
  • amistre64
ok....... I hope your writeing this stuff down :) I might forget it all by tomorrow :)
anonymous
  • anonymous
Explain the four-steps for solving quadratic equations. Can any of these steps be eliminated? Can the order of these steps be changed? Would you add any steps to make it easier, or to make it easier to understand?
anonymous
  • anonymous
No kidding.......writing away!
amistre64
  • amistre64
ive seen this one before in here and it seems to be an odd question in general. the four steps? they might be talking about the 4 common ways that quads are facttored, but then it says to elinate one of them as tho they were all used together.....
amistre64
  • amistre64
the steps I use are: 1) look at the problem 2) factor as needed 3) write down the answer 4) go to next problem?
anonymous
  • anonymous
factor left side? does that make sense?
amistre64
  • amistre64
doesnt clarify my quandry.....no.....
amistre64
  • amistre64
id have to look up and google it to see what its talking about to give a good answer
anonymous
  • anonymous
Hmmm..........ok.
amistre64
  • amistre64
Maybe something like this: Step 1: Find two numbers that add together to get the middle term, and multiply to get the last term. Step 2: Rewrite the equation with those numbers: Step 3: Factor the first two and last two terms separately: Step 4: If you've done this correctly, your two new terms should have a clearly visible common factor.
anonymous
  • anonymous
4) Go to the next problem? HA!
amistre64
  • amistre64
:) yeah. there are many techniques used to factor, I have my favorites, but really, the steps you are looking for will be defined by the source you are using
anonymous
  • anonymous
Can any be elimianted?
anonymous
  • anonymous
OIC.
anonymous
  • anonymous
*eliminated
amistre64
  • amistre64
im gonna go with a no on that one....eliminating steps just sounds bad...and Ive built alot of stairs in my day :)
anonymous
  • anonymous
Could the steps be rearranged?
amistre64
  • amistre64
gonna have to go with a no again for that one, the steps tend to be in order of what you do to get to the next one.....still sounds like a bad idea to be rearranging them.
anonymous
  • anonymous
Lol, ok....well you have been an AWESOME help up until this question anyway! Thanks so much. My brain is cramping up at this point. :)
amistre64
  • amistre64
Ciao bella :)
anonymous
  • anonymous
Lol, I may just write down "it sounds like a bad idea to me" just to see what kind of feedback i get!.
amistre64
  • amistre64
..works for me :)

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