A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 one year ago
Hey, guys here is a summary on algebra by me & give ur opinions on that :)
 one year ago
Hey, guys here is a summary on algebra by me & give ur opinions on that :)

This Question is Closed

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

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

TheViper
 one year 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 \(\Huge{\color{green}{Medal}}\). \(\Huge{\color{orange}{\ddot{\smile}}}\)

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.29Thanx dudes :)

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

shubhamsrg
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0This is not algebra. It's intermediate algebra to be specific!

UnkleRhaukus
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.29@ajprincess @Callisto @shadowfiend @yahhave a look :)

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

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

ghazi
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.29my standard is class 8 only :)

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

whpalmer4
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0Great work:) @jiteshmeghwal9

jiteshmeghwal9
 one year 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
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0Very well done! I look forward to using your textbook when it gets published!

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.29Haha, thanx :)

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

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

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

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

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.29Thanx @Preetha mam :)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.