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jiteshmeghwal9
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Hey, guys here is a summary on algebra by me & give ur opinions on that :)
 one year ago
 one year ago
jiteshmeghwal9 Group Title
Hey, guys here is a summary on algebra by me & give ur opinions on that :)
 one year ago
 one year ago

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jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
\[\Huge{\color{red}{\text{Algebra: Summary}}}\]\[\LARGE{\color{gold}{\star}\color{green}{\text{Quadratic Equation}}}\]The general form of a quadratic equation is \(\color{blue}{ax^2+bx+c=0}\),where \(\color{blue}{a\ne0}\) & \(a, b, c\) are constants.\[\LARGE{\color{green}{\text{solution of quadratic equation}}}\]There are three methods to solve a quadratic equation: 1. Solving by factoring. 2. Solving by completing the square. 3. Solving by Quadratic formula. \[\LARGE{\color{purple}{\text{Solving by factoring:}}}\] For solving by factoring, we must use the zero factor principal. Zero factor principal means in a quadratic equation, one side must be left with a zero i.e; \(ax^2+bx+c=0\) so that we can factor left side & can find the roots of the equation. \[\LARGE{\color{purple}{\text{Solving by completing the square:}}}\] Suppose we have an equation \(ax^2+bx+c=0\), where \(a\ne0\) & a, b, c are constants. Step I Dividing by the coefficient of \(a\) both sides.\[\dfrac{ax^2+bx+c}{a}=\dfrac{0}{a}\]\[x^2+\dfrac{bx}{a}+\dfrac{c}{a}=0.\]Step II Transferring \(\dfrac{c}{a}\) to R.H.S.\[x^2+\dfrac{bx}{a}=\dfrac{c}{a}\]Step III Completing the square now\[x^2+\left( \dfrac{b}{a} \right)x+\left( \dfrac{b}{2a} \right)^2=\left( \dfrac{b}{2a} \right)^2+\dfrac{c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}\dfrac{c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{ab^24a^2c}{4a^3}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{\cancel a(b^24ac)}{\cancel a(4a^2)}\]\[x+\dfrac{b}{2a}=\sqrt{\dfrac{b^24ac}{4a^2}}\]\[x+\dfrac{b}{2a}={\sqrt{b^24ac} \over 2a} \]\[x=\dfrac{b \pm \sqrt{b^24ac}}{2a}\]
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
\[\Large{\color{purple}{\text{Solving by quadratic formula:}}}\]This is a very easy method to solve a quadratic equation. Assume u have an equation x^2+2x+4=0. Here, a=1, b=2 & c=4. Therefore by putting this values in the quadratic formula u can easily get \(x\). Note: Since the quadratic equation's highest degree is 2 therefore its solution also comes in two forms. The main & important part of a quadratic formula is \(\sqrt{b^24ac}=D\) which is called discriminant. It helps us to find which type of solution is this. It has the folloing cases: 1. If D>0, then there are two distinct solutions\[\alpha={b + \sqrt{b^24ac} \over 2a}\]\[\beta=\dfrac{b  \sqrt{b^24ac}}{2a}\]2. If D=0, then the roots are real & equal.\[\alpha={b+0\over2a}=\dfrac{b}{2a}\]\[\beta={b0\over2a}={b \over 2a}\]3. If D<0 then the roots are imaginary\[\alpha=\dfrac{b+\sqrt{1}}{2a}={b+\iota \over 2a}\]\[\beta={b\sqrt{1} \over 2a}=\dfrac{b\iota}{2a}\]
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
\[\LARGE{\color{violet}{\text{Linear Equation In two variables}}}\]An equation in the form of \(ax+by+c=0\), \(ax+by+d=0\) where a & b doesn't equal to zero is called a linear equation in two variables. The values of \(x\) & \(y\) satisfying the equation are called the solutions to it. \[\LARGE{\color{violet}{\text{Simultaneous Linear Equations}}}\]A pair of equations in two variables is called simultaneous linear equations in two variables & the values of \(x\) & \(y\) satisfying the equations are called the solutions to it. General form : \(a_1+b_1=c_1\) \(a_2+b_2=c_2\) (i) \(\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}\) This system has a unique solution. (ii) \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}\) This system has no common solution. (iii) \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\) This system has infinite number of solutions.
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
\[\LARGE{\color{blue}{\text{Division Of Two Polynomials:}}}\] when we divide one polynomial by another polynomial, we got a remainder equals to zero or a number less than the divisor. \(f(x)=g(x).q(x)+r(x)\) where \(g(x)\ne0\) Factor theorem : If f(x) be a polynomial and 'a' be a real number then (xa) is a factor of f(x) if f(a)=0. Remainder theorem : If a polynomial f(x) is divisible by (xa) then the reminder is f(x) where 'a' is a real number.
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@ParthKohli @experimentX @sauravshakya @hartnn @UnkleRhaukus @satellite73 @myininaya @AccessDenied @radar @VincentLyon.Fr @mathslover have a look :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@precal @hba @AravindG have a look :)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
good work @jiteshmeghwal9
 one year ago

TheViper Group TitleBest ResponseYou've already chosen the best response.0
Nice \(\Huge{\color{red}{LaTeX}}\). & the most good thing is that you did so much hard work only to \(teach\). So, you deserved a \(\Huge{\color{green}{Medal}}\). \(\Huge{\color{orange}{\ddot{\smile}}}\)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Thanx dudes :)
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
gr8.....work @jiteshmeghwal9
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
there were just 2 silly mistakes from my part.. 1) when you take sqrt of both sides, + sign comes then only, and not when you finally transfer b/2a to the other side in solving by completing the square you must be knowing this, i'd say just missed that on typing 2)D is not equal to sqrt(b^2 4ac) but is b^2 4ac simply.. otherwise, you've done very good work and i really appreciate your effort ! :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
This is not algebra. It's intermediate algebra to be specific!
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
Great work @jiteshmeghwal9 ! as shubhamsrg pointed out but you should have the discriminant without the square root (I would use a delta instead of a d) \[\Delta=b^24ac\]
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Thanx & i appreciate ur comments & i will improve this mistakes. Thanx to all of u :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@ajprincess @Callisto @shadowfiend @yahhave a look :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@Yahoo!
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@ghazi @ganeshie8 :)
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
this is a good work but i would recommend you to expand it a bit by that i mean to include graph of a quadratic equation and show the solution by curves that will be very helpful for beginners .
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
But it will be expanded to calculus & i do not know calculus :(
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
hmm no i dont think so, but yea its alright :) i thought you have studied calculus :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
my standard is class 8 only :)
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
then its great :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
:)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@Preetha mam have a look :)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
You are using "principal" where "principle" should be used. The principle is: If you choose the wrong word, you have to go to the principal's office :)
 one year ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.0
Great work:) @jiteshmeghwal9
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Ahh..right, lol ;) It is my first big tutorial so little mistakes doesn't matter to me :) Thanx @ajprincess :)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
And you've come across another secret: if you want to understand something really well, try to teach it to someone else!
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
:)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
"Remainder theorem : If a polynomial f(x) is divisible by (xa) then the reminder is f(x) where 'a' is a real number." Shouldn't that be the remainder is f(a)?
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
:P sorry, i meant that the remainder when f(x) is divided by (xa) is f(a) where 'a' is a real number
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Ok ! guys now i'm improving my mistakes. \(\color{red}{\text{*Principal=Principle}}\) \(\color{red}{*\sqrt{b^24ac}=b^24ac}\) \(\color{red}{\text{*Discriminant}=\delta}\) \(\color{red}{\text{*remainder}=f(a), \space \text{not} f(x)}\)
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.0
Very well done! I look forward to using your textbook when it gets published!
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Haha, thanx :)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
capital Delta \Delta
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Okk !
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@TuringTest have a look :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@amistre64 have a look :)
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
@eliassaab @Directrix @Diyadiya
 one year ago

Preetha Group TitleBest ResponseYou've already chosen the best response.0
Good work Jitesh!
 one year ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.29
Thanx @Preetha mam :)
 one year ago
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