anonymous
  • anonymous
can someone explain the quadratic formula to me please
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
Yes, the quadratic formula is what you end up with when you "complete the square". "Completing the square" is proof that the quad formula works. Anything specific you wanna know?
anonymous
  • anonymous
\[5x ^{2} +3x+ 7= 0\]how would i do it with the equation
amistre64
  • amistre64
Just curious, do you know what "quadratic" means? at first I thought it had to do with "4" ... but it is just a fancy name for a squared function like x^2.

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anonymous
  • anonymous
yea im learning about it in math but hes confusing me
amistre64
  • amistre64
Lets start with the basics and then I will get to your equation ok? Tell me what the quad formula is so I know you have a grasp on it.
anonymous
  • anonymous
how to find continuity of function of 2 vars??
amistre64
  • amistre64
Hi think, that doesnt ring a bell. sorry... Are the variables independant or does one depend on the other?
anonymous
  • anonymous
\[-b \sqrt{b ^{2}- 4 (bc} \div 2 (a) \]
anonymous
  • anonymous
its like, lim(x,y)->(1,2){x/sqrt[2x+y]}}
anonymous
  • anonymous
that just confused me
anonymous
  • anonymous
david, for your question, a quadratic equation is which has x power 2, and what you mentioned now is the 'discriminant'
amistre64
  • amistre64
@think: that doesnt look familiar to me. David; it looks like you know the formula; but might have troubles with formatting it with the website... But yes, you are basically right. lets define a, b, and c. a = the first term; b= the middle term, and c= the last term. Does that make sense?
anonymous
  • anonymous
that is a way to find out the roots
anonymous
  • anonymous
yeah after that im lost and i asked him to help me and he confused me even more
anonymous
  • anonymous
@amistre, its like limit of (x,y)->(1,2) with the function as x/ sqrt(2x + y)
amistre64
  • amistre64
lets re-write the top like this: -(middle) +-sqrt(middle^2 - (4)(first)(last)) How does that look to you?
anonymous
  • anonymous
that actually makes sense
amistre64
  • amistre64
@think, I would assume that any values of (x,y) that make the bottom a zero have something to do with it. But I am just not sure.
amistre64
  • amistre64
@David: good then lets call the bottom part of the formula: 2(first) does that help?
anonymous
  • anonymous
yeah
amistre64
  • amistre64
So what was your original equation again and lets see if we can work it out now.
anonymous
  • anonymous
i think twas 5x^2 + 3x + 7 =0
anonymous
  • anonymous
right?
anonymous
  • anonymous
yea
amistre64
  • amistre64
@thnk: what happens when you plug in (1,2) into the equation, does it get a number or an indeterminiate value?
anonymous
  • anonymous
thinker wat grade r u in
amistre64
  • amistre64
5x^2 + 3x + 7 first = (+5) ; middle = (+3); and last = (+7) top of quad form: -3 +- sqrt(3^2 -(4)(5)(7)) ------------------------ bottom part: 2(5) What do we get?
anonymous
  • anonymous
10
anonymous
  • anonymous
sry..it freezed again....i get the answer as 1/2
amistre64
  • amistre64
10 is the bottom of it; but what is the top part of the quad form equal; remember it is just a big fraction looking monstorcity :)
amistre64
  • amistre64
then as (x,y) approaches (1,2) that was the original right? then the equation approaches (1/2) if you plugged it in correctly :)
anonymous
  • anonymous
-3 +- sqrt 9 -140
amistre64
  • amistre64
good; does (-3 + sqrt(-131)) / 10 have any meaning to you? In other words can we have any real values for sqrt(-131)? Let me ask it this way.... what number when multiplied by itself will give you a "negative" answer?+
anonymous
  • anonymous
yeah...so what 's the continuity?
anonymous
  • anonymous
no it has no real solution i think
amistre64
  • amistre64
@think; rationalize the denominator and see if that helps :)
amistre64
  • amistre64
@David; very good, that is the correct answer, there are no "real" values of 'x'. Tell me, what does it mean to be a "real" value?
anonymous
  • anonymous
when a number has a square root??
amistre64
  • amistre64
@ david, good enough answer :) Have you dealt with "imaginary" numbers yet?
anonymous
  • anonymous
no
anonymous
  • anonymous
@amistre, have u anytime seen "Thomas' Calculus" text?
anonymous
  • anonymous
oh r u tlking about the imaginery 1 in front of x
anonymous
  • anonymous
@david, no..imaginary numbers are those which do not exist..like sqrt of a negative number etc
anonymous
  • anonymous
oh
amistre64
  • amistre64
@think; doesnt ring a bell :) @david; lol....good attempt at, but no. the "1" infront of an x is actually there, we just dont write it because it is
amistre64
  • amistre64
1(x) = x the only imaginary number there is, is called "i" and we simply define it as : i = sqrt(-1) it is the only way in which we can solve the squareroots of negative numbers.
anonymous
  • anonymous
ohh..was just a try..not an "algebra" person
anonymous
  • anonymous
@amistre, have u anytime seen "Thomas' Calculus" text??
amistre64
  • amistre64
Thomas Calculus text doesnt sound familiar to me...
anonymous
  • anonymous
ohh..:(..its a huge book..which i have to study from..
amistre64
  • amistre64
I have a few good books in the college library (public libraries here are a joke) that I use. Some are more helpful than others at times.
amistre64
  • amistre64
How did we do with your 1/sqrt(2x+y) question so far?
anonymous
  • anonymous
this function is continous only when the denominator,i.e x,y\[\neq\]0
amistre64
  • amistre64
And do we know if y is a function of x; or if x is a function of y; or if they are independant variables?
anonymous
  • anonymous
there's nothing mentioned...its just that we ve to find the points of continuity/discontinuity
amistre64
  • amistre64
So when (2x+y) = 0 there is a Vertical Asymptote right? When y = -2x and when x = -y/2
anonymous
  • anonymous
vertical asymptote? what is that?
amistre64
  • amistre64
there are a few types of discontinuities available to us: the "hole" is a line is called a jump disconiuity becasue the line will forever get close to it and never touch it. the "vertical" asymptote is a value for "x" that the graph will forever slide right up next to but never touch so instead of a jump across a hole, we get a neverending curve that approaches "x"
anonymous
  • anonymous
sry..didn't get that:(
amistre64
  • amistre64
Holes occur when you can cross out like factors top to bottom. VAs occur when whats left after crossing out makes the bottom equal to "0"
anonymous
  • anonymous
@amistres64 may u pls help me
amistre64
  • amistre64
that was odd... :) kabelo, what is your question? and maybe I can help
amistre64
  • amistre64
@think: think of the graph for f(x) = x^2; and restrict the domain to x<4 and x>4. there is no value for f(x) at x=4; it just makes a hole.
anonymous
  • anonymous
i know is bascially algebra
amistre64
  • amistre64
kab: basic algebra is good :) what is your question?
anonymous
  • anonymous
the veranda is covered with tiles(30cm times 30cm) in 5 black and4 white tiles,how many black tiles are used to cover the veranda floor if that pattern is continued.the area is 4,2m
amistre64
  • amistre64
Kab: Ill go over to that posting , this ones getting rather long ok?
anonymous
  • anonymous
can i join?
anonymous
  • anonymous
yip pls do ,yes u can.
anonymous
  • anonymous
what was the question there? unable to find it:(
anonymous
  • anonymous
the veranda is covered with tiles(30cm times 30cm) in 5 black and4 white tiles,how many black tiles are used to cover the veranda floor if that pattern is continued.the area is 4,2m

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