## jiteshmeghwal9 2 years ago Hey, guys here is a summary on algebra by me & give ur opinions on that :)

1. jiteshmeghwal9

$\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}$

2. jiteshmeghwal9

$\Large{\color{purple}{\text{Solving by quadratic formula:-}}}$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}$

3. jiteshmeghwal9

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

4. jiteshmeghwal9

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

5. jiteshmeghwal9

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

6. jiteshmeghwal9

@precal @hba @AravindG have a look :)

7. AravindG

good work @jiteshmeghwal9

8. TheViper

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

9. jiteshmeghwal9

Thanx dudes :)

10. mayankdevnani

gr8.....work @jiteshmeghwal9

11. shubhamsrg

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 ! :)

12. ParthKohli

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

13. UnkleRhaukus

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) $\Delta=b^2-4ac$

14. jiteshmeghwal9

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

15. jiteshmeghwal9

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

16. jiteshmeghwal9

@Yahoo!

17. jiteshmeghwal9

@ghazi @ganeshie8 :)

18. ghazi

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 .

19. jiteshmeghwal9

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

20. ghazi

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

21. jiteshmeghwal9

my standard is class 8 only :)

22. ghazi

then its great :)

23. jiteshmeghwal9

:)

24. jiteshmeghwal9

@Preetha mam have a look :)

25. whpalmer4

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 :-)

26. ajprincess

Great work:) @jiteshmeghwal9

27. jiteshmeghwal9

Ahh..right, lol ;) It is my first big tutorial so little mistakes doesn't matter to me :) Thanx @ajprincess :)

28. whpalmer4

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

29. jiteshmeghwal9

:)

30. whpalmer4

"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)?

31. jiteshmeghwal9

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

32. jiteshmeghwal9

Ok ! guys now i'm improving my mistakes. $$\color{red}{\text{*Principal=Principle}}$$ $$\color{red}{*\sqrt{b^2-4ac}=b^2-4ac}$$ $$\color{red}{\text{*Discriminant}=\delta}$$ $$\color{red}{\text{*remainder}=f(a), \space \text{not} f(x)}$$

33. LogicalApple

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

34. jiteshmeghwal9

Haha, thanx :)

35. UnkleRhaukus

capital Delta \Delta

36. jiteshmeghwal9

Okk !

37. jiteshmeghwal9

@TuringTest have a look :)

38. jiteshmeghwal9

@amistre64 have a look :)

39. jiteshmeghwal9

40. Preetha

Good work Jitesh!

41. jiteshmeghwal9

Thanx @Preetha mam :)