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

- jiteshmeghwal9

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

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

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

- 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\[\alpha=\dfrac{-b+\sqrt{-1}}{2a}={-b+\iota \over 2a}\]\[\beta={-b-\sqrt{-1} \over 2a}=\dfrac{b-\iota}{2a}\]

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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.

- jiteshmeghwal9

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

- AravindG

good work @jiteshmeghwal9

- 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 \(\Huge{\color{green}{Medal}}\).
\(\Huge{\color{orange}{\ddot{\smile}}}\)

- jiteshmeghwal9

Thanx dudes :)

- mayankdevnani

gr8.....work @jiteshmeghwal9

- 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 ! :)

- ParthKohli

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

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

- jiteshmeghwal9

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

- jiteshmeghwal9

- jiteshmeghwal9

- jiteshmeghwal9

@ghazi @ganeshie8 :)

- 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 .

- jiteshmeghwal9

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

- ghazi

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

- jiteshmeghwal9

my standard is class 8 only :)

- ghazi

then its great :)

- jiteshmeghwal9

:)

- jiteshmeghwal9

@Preetha mam have a look :)

- 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 :-)

- ajprincess

Great work:) @jiteshmeghwal9

- jiteshmeghwal9

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

- whpalmer4

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

- jiteshmeghwal9

:)

- 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)?

- 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

- jiteshmeghwal9

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

- anonymous

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

- jiteshmeghwal9

Haha, thanx :)

- UnkleRhaukus

capital Delta \Delta

- jiteshmeghwal9

Okk !

- jiteshmeghwal9

@TuringTest have a look :)

- jiteshmeghwal9

@amistre64 have a look :)

- jiteshmeghwal9

- Preetha

Good work Jitesh!

- jiteshmeghwal9

Thanx @Preetha mam :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.