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Today, My teacher was sort of, tricked us and used confusing explanation on "Solving Quadratic Equations by Factoring"
Then I got a Score of 2/10, The Answers I saw was like, where did he even get this and he even didn't even tell us.
So can anyone help me to try to do this?
Confusing Problem
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Ill send the full question later
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The general rule for algebra is that you can do anything you can do to normal numbers as long as you do them to both sides, so if you had x = y then you could multiply both sides by 2 to get 2x = 2y, or subtract 2 to get x - 2 = y - 2.
In this case you are subtracting one from both sides, which gives you 2x + 1 - 1 = 0 - 1, which you can then simplify to get 2x = -1, effectively 'moving' the 1 from one side to the other.
You can then just divide by two to get 2x / 2 = -1 / 2, which gives x = -1/2.
Hope this helps
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For x = y, x and y in set of real numbers you can get ...
ax = ay
x + a = y + a
x^a = y^a
a^x = a^y
... for any a in set of real numbers. That means that you can:
Add and subtract one number from both sides of an equation.
Multiply and divide by any number not zero from both sides of an equation.
Take any power or root from both sides of an equation.
Edit:
2x + 1 = 0
2x + 1 + (-1) = -1
2x = -1
2x * (1/2) = -1 * (1/2)
(2x/2) = -1/2
x = - 1/2
What that’s easy.
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Heres the Full question
4x^2 + 4x + 1 = 0
Thanks for the answer
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Heres the Full question
4x^2 + 4x + 1 = 0Thanks for the answer
4x^2 + 4x + 1 = (2x)^2 + 2 * (2x) * (1) + (1) = (2x+1)^2
(2x+1)^2 = 0
(2x+1)^2^(1/2) = 0^(1/2)
|2x+1| = 0
Heres the Full question
4x^2 + 4x + 1 = 0Thanks for the answer
You have to know the formula for quadratic equations
That is, for the equation
ax^2 + bx + c = 0
x = (-b +- sqrt(b^2 - 4ac))/2a
So in your equation
x = (-4 +- sqrt(16 - 4*4*1))/8
x = (-4 +- sqrt(0))/8
x = -4/8
x = -1/2
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what i was taught was:
5x - 4 = 2x + 8
take away the small x thing from the big x thing
3x - 4 = 8
then do stuff
3x = 8 + 4
then you divide the answer to the 8+3 by the number x thing
x = 12/3
tada you done fam
x = 4
Heres the Full question
4x^2 + 4x + 1 = 0Thanks for the answer
i would use the metoh for 'guessing the answer' guess an answer you think is right and fill it in the X's try until you found the answe is what works with an ^ sign otherwise lukems thing works in your first question whic is ez
thanks hg for making this much better and ty for my avatar aswell
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Heres the Full question
4x^2 + 4x + 1 = 0Thanks for the answer
This thing, how is it called
I know you use this for those questions
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AllenCaspe9510 wrote:Heres the Full question
4x^2 + 4x + 1 = 0Thanks for the answer
This thing, how is it called
https://imgur.com/wSRGUj2.png
I know you use this for those questions
Uhh... No, I might see it sometime
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what i was taught was:
5x - 4 = 2x + 8
take away the small x thing from the big x thing
3x - 4 = 8
then do stuff
3x = 8 + 4
then you divide the answer to the 8+3 by the number x thing
x = 12/4
tada you done fam
x = 3
Why 3? Why do you divide by 4? X = 12/3 -> x = 4
MWstudios wrote:AllenCaspe9510 wrote:Heres the Full question
4x^2 + 4x + 1 = 0Thanks for the answer
This thing, how is it called
https://imgur.com/wSRGUj2.png
I know you use this for those questionsUhh... No, I might see it sometime
It works like this (or at least how I remember it):
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4x^2 + 4x + 1 = 0
This thing is called a quadratic equation. Quadratic equations are base 2 (meaning the leading factor is to the power of two). They're written like this... ax^2 + bx + c = 0.
In order to solve these, there are a few methods, but the most reliable way to solve these is by using the quadratic formula... the thing MWStudios posted earlier...
In order to use the quadratic formula, just "plug in" the numbers from the original equation into where it belongs on the formula, and solve it.
ax^2 + bx + c = 0.
This is how they are written
4x^2 + 4x + 1 = 0
This is the equation you're solving... a=4, b=4, and c=1, so plug those numbers into the equation and solve.
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Just FYI the OP's question specifically said you had to solve it by factoring, so you can't just throw the formula at it and write down what you get, you need to try to get it to the form (ax + b)(cx + d) = 0, usually just by inspection, then you can say that either side must be 0, so x = -b/a or -d/c.
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sxrrealism wrote:what i was taught was:
5x - 4 = 2x + 8
take away the small x thing from the big x thing
3x - 4 = 8
then do stuff
3x = 8 + 4
then you divide the answer to the 8+3 by the number x thing
x = 12/4
tada you done fam
x = 3Why 3? Why do you divide by 4? X = 12/3 -> x = 4
ok sorry i did a dumb thing i made a typo
what i was taught was:
5x - 4 = 2x + 8
take away the small x thing from the big x thing
3x - 4 = 8
then do stuff
3x = 8 + 4
then you divide the answer to the 8+3 by the number x thing
x = 12/3
tada you done fam
x = 4
what i was though was
5x-4=2x+8
3x-4=8
3x=12
12/3x=4
x=4
or
5x-4=2x+8
-4=-3x+8
-12=-3x
-12/-3x=4
x=4
basically get all the variables in this cases the X's on one side by adding/subtractign them evenly (check if positive or negative numbers!) then get all looose nubmbers on otther side then its an easy division (or not) tehn you COULD checkyourself by putting you answer onthe X's and get 5x-4=2x+8 5*4-4=16 2*4+8=16 = even its just like a weightscale thaking blocks form 1KG for loose numbers and like other blocks or objects for the X take away even from bto sdies and tis still even (doesntwork when negative numbers but its still same ;p)
thanks hg for making this much better and ty for my avatar aswell
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How about solving Quadratic Equations by Completing the square for binomial and trinomial?
I was also confused about those
Three of this is quite annoying to understand
[] (h - 3/5)^2 = 1/2
[] s^2 + 3/4 s + 9/64
[] h^2 - 3/4 h + ___?
Edit:
I'm not doing this for my assignment, I need a formula :/
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ax^2 + bx + c
= a(x^2 + bx/a) + c
= a(x^2 + bx/a + (b/2a)^2) - b^2/4a + c
= a(x + (b/2a))^2 - b^2/4a + c
Which gives you a quadratic equation in vertex form (y=a(x-h)^2+k, where h = -b/2a, k = -b^2/4a+c)
notes:
a(b/2a)^2 = a(b^2/4a^2) = b^2/4a
x^2 + 2xy + y^2 = (x+y)^2 (where y = b/2a)
If you only care about your final answer (your math teacher is bad if this is the case), you only really need to know the last line in that equation.
This was a nice refresher. I haven't needed to do basic math like this for about 5 months.
If you need more help, I'd recommend Khan Academy.
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To help you visualize math use https://www.desmos.com/calculator
if you want to check your answer, plug in the formula and you'll get a visualized result
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To help you visualize math use https://www.desmos.com/calculator
if you want to check your answer, plug in the formula and you'll get a visualized result
https://gosha.eejesse.net/files/7910468d.png
nice calcultor
thanks hg for making this much better and ty for my avatar aswell
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To help you visualize math use https://www.desmos.com/calculator
if you want to check your answer, plug in the formula and you'll get a visualized result
https://gosha.eejesse.net/files/7910468d.png
Is that a Geogebra bootleg?
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