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 2 years ago
Hey, guys here is a summary on algebra by me & give ur opinions on that :)
 2 years ago
Hey, guys here is a summary on algebra by me & give ur opinions on that :)

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jiteshmeghwal9
 2 years ago
Best 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\[x^2+\left( \dfrac{b}{a} \right)x+\left( \dfrac{b}{2a} \right)^2=\left( \dfrac{b}{2a} \right)^2+\dfrac{c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}\dfrac{c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{ab^24a^2c}{4a^3}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{\cancel a(b^24ac)}{\cancel a(4a^2)}\]\[x+\dfrac{b}{2a}=\sqrt{\dfrac{b^24ac}{4a^2}}\]\[x+\dfrac{b}{2a}={\sqrt{b^24ac} \over 2a} \]\[x=\dfrac{b \pm \sqrt{b^24ac}}{2a}\]

jiteshmeghwal9
 2 years ago
Best 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\[\alpha=\dfrac{b+\sqrt{1}}{2a}={b+\iota \over 2a}\]\[\beta={b\sqrt{1} \over 2a}=\dfrac{b\iota}{2a}\]

jiteshmeghwal9
 2 years ago
Best 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.

jiteshmeghwal9
 2 years ago
Best 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.

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

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@precal @hba @AravindG have a look :)

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0good work @jiteshmeghwal9

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

mayankdevnani
 2 years ago
Best ResponseYou've already chosen the best response.0gr8.....work @jiteshmeghwal9

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1there 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 ! :)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0This is not algebra. It's intermediate algebra to be specific!

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0Great 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) \[\Delta=b^24ac\]

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

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@ajprincess @Callisto @shadowfiend @yahhave a look :)

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@ghazi @ganeshie8 :)

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.1this 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 .

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29But it will be expanded to calculus & i do not know calculus :(

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

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29my standard is class 8 only :)

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@Preetha mam have a look :)

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.1You 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 :)

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0Great work:) @jiteshmeghwal9

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

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

whpalmer4
 2 years ago
Best 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)?

jiteshmeghwal9
 2 years ago
Best 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

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

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0Very well done! I look forward to using your textbook when it gets published!

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0capital Delta \Delta

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@TuringTest have a look :)

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@amistre64 have a look :)

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29@eliassaab @Directrix @Diyadiya

jiteshmeghwal9
 2 years ago
Best ResponseYou've already chosen the best response.29Thanx @Preetha mam :)
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