Question

Can anyone explain in great detail what the Quadractic Formula means and how I can solve Quadratic Equations but for a dummy like me

Answer 1

I will give medal to first person to explain it to me in a way that I can understand it

Answer 2

If you wanna know why I am asking these questions it is because I might not get into the schools I am applying to for 6th grade

Answer 3

even though I got a decent score of about 1900 on the middle school ssat

Answer 4

I believe I got an 85th percentile on the ssat

Answer 5

@Mehek14 @pooja195

Answer 6

@tkhunny

Answer 7

@SolomonZelman

Answer 8

pretty sad that I might not get in

Answer 9

to the few schools I am applying to

Answer 10

The general quadratic formula is used to find the values of x that make the quadratic equation equal zero.

There are the points where the parabola cuts of touches the x- axis.

All parabola's are symmetric

so given a quadratic equation.

\[y = ax^2 + bx + c\]

the line of symmetry is found using

\[x = \frac{-b}{2a}\]

so the points that are equi-distance from the line of symmetry can be found using the general quadratic formula.

to find the point(s) where the parabola cuts the x-axis set y = 0 so then you have

\[0 = ax^2 + bx + c\]

the general quadratic formula is

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

you may notice that the line of symmetry equation is part of the quadratic formula..

\[\pm \sqrt{b^2 - 4ac}\]

is used to calculate a distance from the line of symmetry on both sides of the line.

hope it makes sense.

Answer 11

so that is why I am learning all this

Answer 12

can you tell me what you just said but in fewer words and in an easier way to think of it

Answer 13

wow, do you save topic answers or just type all tha tnow.

Answer 14

I just typed it off the top of my head @DanJS

Answer 15

cause I am in middle school and not in 7th grade yet but wanna learn it since I might not get into my dream schools

Answer 16

general quadratic, a,b,c are constant numbers

y = ax^2+bx+c

when y=0, the graph crosses the x axis

0 = ax^2 + bx + c

the formula is just solving that there for x, nice to show it, or people just remember

Answer 17

is ^ used for exponents?

Answer 18

yes, sorry, good for you learning algebra in 7th grade, dont think i had that till 8th

Answer 19

also I am not in 7th grade

Answer 20

I just said my sister learned it in her class for 8th grade or 7th grade so I just wanna try and learn this before I get into her grade to try again at the SSAT and possibly do better

Answer 21

cause I did get like an 85th percentile on the ssat but no schools look like taking me even though I didn't get any letters yet

Answer 22

cause I think they just don't like me

Answer 23

also can you show me on a graph what it would look like

Answer 24

so I can get a better visual idea of it

Answer 25

@satellite73

Answer 26

the quadratic form has highest power x^2, these are parabolas, sometimes there might not be any real number solutions to the formula or xintercepts

Answer 27

I gtg

Answer 28

bye

Answer 29

I am back

Answer 30

@jhonyy9 @triciaal @SolomonZelman @kropot72

Answer 31

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjMj8ndw9_KAhXFlR4KHTPfALAQFggcMAA&url=https%3A%2F%2Fwww.khanacademy.org%2Fmath%2Falgebra%2Fquadratics%2Fsolving-quadratics-using-the-quadratic-formula%2Fa%2Fquadratic-formula-explained&usg=AFQjCNFbsPBnIz1njzkxBHV6QxBQjbke_g&sig2=oduIjmY_vo2IkyeEQQJTTQ

Answer 32

hope this one help

Answer 33

yeah those are easy videos to follow on that site, start the algebra ones

Answer 34

I think I saw something that I can understand https://www.youtube.com/watch?v=70dA34KAFWw

Answer 35

can you check to see if it is legit

Answer 36

ok

Answer 37

is this legit? https://www.youtube.com/watch?v=velArLW85MU

Answer 38

@DanJS

Answer 39

@triciaal

Answer 40

I think I got it now

Answer 41

maybe I can test my skills to see if I know it

Answer 42

i never seen those..

here, play with this site
https://www.khanacademy.org/math/algebra-basics

make an account on that site , free, start wherever you are comfortable, it is organized into subjects

Answer 43

i send him that danJS

Answer 44

@DanJS

Answer 45

could you test me with a problem or two?

Answer 46

so that I know if I do know it or not

Answer 47

oh right, yeah that is in algebra 1 section, you should be comfortable with all the stuff before that first, then move to those

Answer 48

lines,linear equations. and manipulating equations around, and stuff

Answer 49

I don't think I understand it

Answer 50

I really don't understand the quadratic formula
and how to solve something like -4x + 2 -4x = 0

Answer 51

everything I think is just incorrect and I have no idea on how to solve anything

Answer 52

Would you know how to solve something like this? :)
3x-7+4x=0

Answer 53

I have no idea @zepdrix

Answer 54

I really want to learn this

Answer 55

I need to learn as much algebra as I can and geometry since I might have to go back to my old school

Answer 56

since no school seems interested in me

Answer 57

ax^2 + bx + c = 0

Solve by Completing the Square

That's all there is to it.

Answer 58

Well first, make sure you're comfortable with the idea of `like-terms`.
We can combine things together if they are the same in a way.

Example:
3x-7
These are different "things" being subtracted,
they can't be combined in any meaningful way.
But instead if we look at
3x+4x
These are just different amounts of "x", whatever x may be.
Let's call it an apple.
So this example is really 3apples + 4apples, which we know to be 7apples.
So therefore,
3x+4x = 7x

Answer 59

So when we're given a problem like:
3x-7+4x=0

We want to start by combining `like-terms`.
We want to combine anything that can be combined.
So adding the 4x to the 3x gives us this next step:
7x-7=0

Answer 60

3x - 7 + 4x = 0 meaning you would do 3x + 4x = 7 making x = 1

Answer 61

Or add 7 to both sides as you've shown,
yes that works just as well :)

Answer 62

ahh sorry i have to go D:

Answer 63

so would I do this 3x - 7 + 4x = 0 and then add 7 to both sides giving us 3x+ 4x = 7 and then find out what is a b and c which are a = 3 b = 4 and c = -7 then divide b by 2 then square it so 2 x 2 = 4 so now you have 3x + 4x + 4 which is factorable. You can factor 2 and 2 giving you the solution of 2

Answer 64

so yea that is the only solution you would have

Answer 65

If I am right than could you tell me and if I didn't could you show me everything that I did wrong

Answer 66

no, that equation is a line, no x^2 term in there,
start https://www.khanacademy.org/math/algebra-basics
go thorugh foundations and work through, come here to ask questions for examples

Answer 67

i would recommend, prolly

Answer 68

Liam that would work because if you have 3x+4y=7 first the equatoon will have to equal 0 when doing the quadratic formula and you have to have it in the form of x^2+x+c=0

Answer 69

*wouldnt

Answer 70

If you want to we could do anexample

Answer 71

if you have 3x^2-6x=3 so you than you will have to move the 3 to the other side by subtractinf it, so 3x^2-6x-3=0 than you can call a 3 b -6 and c -3 so nuow plug that into the equatoon -b+- b^2-4ac all under the square root divided by 2a

Answer 72

So 6 +- 36+36 so (6+-square root 72)/6

Answer 73

I really don't understand

Answer 74

@dan815

Answer 75

@ganeshie8

Answer 76

@nincompoop @ParthKohli

Answer 77

@whpalmer4

Answer 78

also if you can explain it to me and tell me how to solve it then I will give you a medal and a fan

Answer 79

@hartnn

Answer 80

@jigglypuff314

Answer 81

A quadratic equation is a 2nd-degree equation usually in only one variable. 2nd degree means that the variable is squared. A typical quadratic equation is a three-term, second degree polynomial (also called a quadratic trinomial) equaling zero.
The typical form of a quadratic equation is:

\(ax^2 + bx + c = 0\)

where x is the variable, and a, b, and c are just numbers.

Here are some examples of quadratic equations:
\(x^2 + 2x - 5 = 0\)
\(2x^2 + 6x - 7 = 0\)
\(5v^2 - 2v + 2 = 0\)
\(2s^2 + 9 = 0\)
\(5w^2 =0\)

Notice that all the equations have a term with the variable squared. All quadratic equations are a polynomial of second degree equaling zero.

The goal in solving the quadratic equation is to find what values of the variable will make the polynomial equal zero.

There are several ways of solving a quadratic equation. One such method is called the complete the square method. The quadratic equation is derived from the complete the square method.

Without going through the complete derivation of the quadratic formula, here is the quadratic formula:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In order to use it to solve a quadratic equation, all you need to do is write the quadratic equation is the form

\(ax^2 + bx + c = 0\)

Then use the numbers a, b, and c in the quadratic formula to find the solutions.

Here is an example.

Solve using the quadratic formula:

\(2x^2 - 9x - 18 = 0\)

When you compare this equation with the form

\(ax^2 + bx + c = 0\)

you see that a = 2, b = -9, and c = -18.

Now write the quadratic formula, and then replace a, b, and c with the values from our quadratic equation:

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\(x = \dfrac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(-18)}}{2(2)} \)

\(x = \dfrac{9 \pm \sqrt{81 + 144}}{4} \)

\(x = \dfrac{9 \pm \sqrt{225}}{4} \)

\(x = \dfrac{9 \pm 15}{4} \)

\(x = \dfrac{9 + 15}{4} \) or \(x = \dfrac{9 - 15}{4} \)

\(x = \dfrac{24}{4} \) or \(x = \dfrac{-6}{4} \)

\(x = 6 \) or \(x = -\dfrac{3}{2} \)

If you test the two solutions in the given equation, you will see that those two values of x will indeed make the value of the polynomial zero.

Answer 82

Think of the quadratic formula as a recipe for giving you the solution that always works.

If you can rearrange your formula into the form\[ax^2+bx+c=0\]where \(a \ne 0\) then you can always get the values of \(x\) which make that true by evaluating
\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

There are other ways you can solve a quadratic as well; some are easier to understand or use in some cases.

Factoring:

if you can factor your quadratic into the form \[k(x-a)(x-b) = 0\]where \(k,a,b\) are constants, then your solutions are simply the values of \(x\) that make \(x-a=0\) and \(x-b=0\) true. For example:
\[x^2-3x+2\]can be factored as \[(x-1)(x-2)\]because \[(x-1)(x-2) = x^2-2x-1x+2 = x^2 -3x + 3\]If we set those factors equal to \(0\) and solve for \(x\) we get
\[x-2=0, \ x=2\]\[x-1 = 0,\ x=1\]

And sure enough, look what happens when we plug those values into our equation:
\[x^2-3x+2 = (1)^2-3(1)+2 = 1-3+2 = 0\]\[(2)^2-3(2)+2 = 4-6+2 = 0\]

Answer 83

Completing the square:

Another way is what is called "completing the square". Here we have something like \[x^2 + bx + c = 0\]

Now imagine we have \[(x+\frac{b}{2})^2 = (x+\frac{b}{2})(x+\frac{b}{2}) = x^2 + x\frac{b}{2} + x\frac{b}{2} + (\frac{b}{2})^2 = x^2 + bx + \frac{b^2}{4}\]
That means we could write \(x^2+bx\) as \((x+\frac{b}{2})^2\) as long as we fudge the rest of the equation a bit by simultaneously adding and subtracting \((b/2)^2\). Let me solve the same equation from before:
\[x^2-3x+2 = 0\]

\[(x^2-3x) + 2 = 0\]We take half of the coefficient of \(x\) (in other words, the number in front of \(x\)), square it, and both add and subtract that number. If we both add and subtract some number, there is no change in the value. If you have $20 in your wallet, and I give you a $10 bill and I take a $10 bill from your wallet, you still have $20. We're going to do the same thing.

\[(x^2-3x+(\frac{3}{2})^2) + 2 - (\frac{3}{2})^2 = 0\]\[(x^2-3x+\frac{9}{4})+2-\frac{9}{4} = 0\]The stuff in the parentheses we can rewrite as \((x-\frac{3}{2})^2\)
\[(x-\frac{3}{2})^2 + 2 - \frac{9}4=0\]Move the constants to the other side and combine:
\[(x-\frac{3}{2})^2 = \frac{9}4-2 = \frac{1}{4}\]Now we take the square root of both sides:
\[\pm(x-\frac{3}{2}) = \frac{1}{2}\]
I have the \(\pm\) because both the negative and positive square roots give the same result when squared.

\[+(x-\frac{3}{2}) = \frac{1}2\]\[x=2\]\[-(x-\frac{3}{2}) =\frac{1}{2}\]\[-x=-1\]\[x=1\]

Same solutions as before.

Answer 84

Finally, using the quadratic formula:

\[x^2-3x+2=0\]\[ax^2+bx+c=0\]comparing like terms, we can see that \[a=1,\ b=-3,\ c=2\]so our solutions from the quadratic formula are:
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-(-3)\pm\sqrt{(-3)^2-4(1)(2)}}{2(1)} =\frac{3\pm\sqrt{9-8}}{2} \]\[=\frac{3\pm1}{2} = \frac{4}2,\frac{2}2=2,1 \]

The quadratic formula is especially useful for real-world problems where the coefficients are decimals, maybe something like \[-0.254x^2 +5.9730x - 23.990141 = 0\]where attempting to factor or complete the square would be a big headache, even if you turned those decimals into fractions, such as \[-\frac{254}{1000}x^2+\frac{5973}{1000}-\frac{23990141}{1000000}=0\]

Answer 85

I have no idea still and wish to learn

Answer 86

hehhe

Answer 87

hmm have you covered quadratic equations yet?
I mean... say... can you throw me one, any one, make any up

Answer 88

no but wish to since I am applying to prestigious east coast schools so I am wanting to learn how to do this when I am going into 6th grade

Answer 89

and also I have the option to skip grades in math

Answer 90

so I am trying to learn this one by one

Answer 91

well... if you hmmm haven't covered quadratics yet
this will be a good time to go through them

for that is what the quadratric formula is for
so. you want to cover quadratics firstly

Answer 92

found you!

Answer 93

I know the quadratic formula and how to solve a problem it is just I know some but not others

Answer 94

what were saying about medals

Answer 95

oh that you press best response to give somebody one.

Answer 96

I hope you understand what I mean