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jiteshmeghwal9
Hey, guys here is a summary on algebra by me & give ur opinions on that :)
\[\Huge{\color{red}{\text{Algebra: Summary}}}\]\[\LARGE{\color{gold}{\star}\color{green}{\text{Quadratic Equation}}}\]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\[x^2+\left( \dfrac{b}{a} \right)x+\left( \dfrac{b}{2a} \right)^2=\left( \dfrac{b}{2a} \right)^2+\dfrac{-c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}-\dfrac{c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{ab^2-4a^2c}{4a^3}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{\cancel a(b^2-4ac)}{\cancel a(4a^2)}\]\[x+\dfrac{b}{2a}=\sqrt{\dfrac{b^2-4ac}{4a^2}}\]\[x+\dfrac{b}{2a}={\sqrt{b^2-4ac} \over 2a} \]\[x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[\Large{\color{purple}{\text{Solving by quadratic formula:-}}}\]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 \(\sqrt{b^2-4ac}=D\) 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\[\alpha={-b + \sqrt{b^2-4ac} \over 2a}\]\[\beta=\dfrac{-b - \sqrt{b^2-4ac}}{2a}\]2. If D=0, then the roots are real & equal.\[\alpha={-b+0\over2a}=\dfrac{-b}{2a}\]\[\beta={-b-0\over2a}={-b \over 2a}\]3. If D<0 then the roots are imaginary\[\alpha=\dfrac{-b+\sqrt{-1}}{2a}={-b+\iota \over 2a}\]\[\beta={-b-\sqrt{-1} \over 2a}=\dfrac{b-\iota}{2a}\]
\[\LARGE{\color{violet}{\text{Linear Equation In two variables}}}\]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.
\[\LARGE{\color{blue}{\text{Division Of Two Polynomials:-}}}\] 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 (x-a) is a factor of f(x) if f(a)=0. Remainder theorem :- If a polynomial f(x) is divisible by (x-a) then the reminder is f(x) where 'a' is a real number.
@ParthKohli @experimentX @sauravshakya @hartnn @UnkleRhaukus @satellite73 @myininaya @AccessDenied @radar @Vincent-Lyon.Fr @mathslover have a look :)
@precal @hba @AravindG have a look :)
good work @jiteshmeghwal9
Nice \(\Huge{\color{red}{LaTeX}}\). & the most good thing is that you did so much hard work only to \(teach\). So, you deserved a \(\Huge{\color{green}{Medal}}\). \(\Huge{\color{orange}{\ddot{\smile}}}\)
gr8.....work @jiteshmeghwal9
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 ! :)
This is not algebra. It's intermediate algebra to be specific!
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) \[\Delta=b^2-4ac\]
Thanx & i appreciate ur comments & i will improve this mistakes. Thanx to all of u :)
@ajprincess @Callisto @shadowfiend @yahhave a look :)
@ghazi @ganeshie8 :)
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 .
But it will be expanded to calculus & i do not know calculus :(
hmm no i dont think so, but yea its alright :) i thought you have studied calculus :)
my standard is class 8 only :)
@Preetha mam have a look :)
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 :-)
Great work:) @jiteshmeghwal9
Ahh..right, lol ;) It is my first big tutorial so little mistakes doesn't matter to me :) Thanx @ajprincess :)
And you've come across another secret: if you want to understand something really well, try to teach it to someone else!
"Remainder theorem :- If a polynomial f(x) is divisible by (x-a) then the reminder is f(x) where 'a' is a real number." Shouldn't that be the remainder is f(a)?
:P sorry, i meant that the remainder when f(x) is divided by (x-a) is f(a) where 'a' is a real number
Ok ! guys now i'm improving my mistakes. \(\color{red}{\text{*Principal=Principle}}\) \(\color{red}{*\sqrt{b^2-4ac}=b^2-4ac}\) \(\color{red}{\text{*Discriminant}=\delta}\) \(\color{red}{\text{*remainder}=f(a), \space \text{not} f(x)}\)
Very well done! I look forward to using your textbook when it gets published!
capital Delta \Delta
@TuringTest have a look :)
@amistre64 have a look :)
@eliassaab @Directrix @Diyadiya
Thanx @Preetha mam :)