Question

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

Answer 1

\[\Huge{\color{red}{\text{Algebra: Summary}}}\]\[\LARGE{\color{gold}{\star}\color{green}{\text{Quadratic Equation}}}\]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\[x^2+\left( \dfrac{b}{a} \right)x+\left( \dfrac{b}{2a} \right)^2=\left( \dfrac{b}{2a} \right)^2+\dfrac{-c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}-\dfrac{c}{a}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{ab^2-4a^2c}{4a^3}\]\[\left( x+\dfrac{b}{2a} \right)^2=\dfrac{\cancel a(b^2-4ac)}{\cancel a(4a^2)}\]\[x+\dfrac{b}{2a}=\sqrt{\dfrac{b^2-4ac}{4a^2}}\]\[x+\dfrac{b}{2a}={\sqrt{b^2-4ac} \over 2a} \]\[x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 2

\[\Large{\color{purple}{\text{Solving by quadratic formula:-}}}\]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\[\alpha=\dfrac{-b+\sqrt{-1}}{2a}={-b+\iota \over 2a}\]\[\beta={-b-\sqrt{-1} \over 2a}=\dfrac{b-\iota}{2a}\]

Answer 3

\[\LARGE{\color{violet}{\text{Linear Equation In two variables}}}\]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.

Answer 4

\[\LARGE{\color{blue}{\text{Division Of Two Polynomials:-}}}\] 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.

Answer 5

@ParthKohli @experimentX @sauravshakya @hartnn @UnkleRhaukus @satellite73 @myininaya @AccessDenied @radar @Vincent-Lyon.Fr @mathslover have a look :)

Answer 6

@precal @hba @AravindG have a look :)

Answer 7

good work @jiteshmeghwal9

Answer 8

Nice \(\Huge{\color{red}{LaTeX}}\).
& the most good thing is that you did so much hard work only to \(teach\).
So, you deserved a \(\Huge{\color{green}{Medal}}\).

\(\Huge{\color{orange}{\ddot{\smile}}}\)

Answer 9

Thanx dudes :)

Answer 10

gr8.....work @jiteshmeghwal9

Answer 11

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 ! :)

Answer 12

This is not algebra. It's intermediate algebra to be specific!

Answer 13

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) \[\Delta=b^2-4ac\]

Answer 14

Thanx & i appreciate ur comments & i will improve this mistakes. Thanx to all of u :)

Answer 15

@ajprincess @Callisto @shadowfiend @yahhave a look :)

Answer 16

@Yahoo!

Answer 17

@ghazi @ganeshie8 :)

Answer 18

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 .

Answer 19

But it will be expanded to calculus & i do not know calculus :(

Answer 20

hmm no i dont think so, but yea its alright :) i thought you have studied calculus :)

Answer 21

my standard is class 8 only :)

Answer 22

then its great :)

Answer 23

:)

Answer 24

@Preetha mam have a look :)

Answer 25

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 :-)

Answer 26

Great work:) @jiteshmeghwal9

Answer 27

Ahh..right, lol ;)

It is my first big tutorial so little mistakes doesn't matter to me :)



Thanx @ajprincess :)

Answer 28

And you've come across another secret: if you want to understand something really well, try to teach it to someone else!

Answer 29

:)

Answer 30

"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)?

Answer 31

:P sorry, i meant that the remainder when f(x) is divided by (x-a) is f(a) where 'a' is a real number

Answer 32

Ok ! guys now i'm improving my mistakes.

\(\color{red}{\text{*Principal=Principle}}\)
\(\color{red}{*\sqrt{b^2-4ac}=b^2-4ac}\)
\(\color{red}{\text{*Discriminant}=\delta}\)
\(\color{red}{\text{*remainder}=f(a), \space \text{not} f(x)}\)

Answer 33

Very well done! I look forward to using your textbook when it gets published!

Answer 34

Haha, thanx :)

Answer 35

capital Delta \Delta

Answer 36

Okk !

Answer 37

@TuringTest have a look :)

Answer 38

@amistre64 have a look :)

Answer 39

@eliassaab @Directrix @Diyadiya

Answer 40

Good work Jitesh!

Answer 41

Thanx @Preetha mam :)