A community for students.
Here's the question you clicked on:
 0 viewing
klimenkov
 3 years ago
Tutorial. How to solve the quadratic equation.
klimenkov
 3 years ago
Tutorial. How to solve the quadratic equation.

This Question is Closed

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.9Any number in this article is REAL. Definition: The quadratic equation is an expression:\[ax^2+bx+c=0\]where \(a,b,c \) are any numbers, but \(a\ne0\). 1. There are 2 solutions of this equation in case if \(b^24ac>0\) and they can be found in this way:\[x_1=\frac{b\sqrt{b^24ac}}{2a}, \ x_2=\frac{b+\sqrt{b^24ac}}{2a}\]2. There is 1 solution of this equation in case if \(b^24ac=0\) and it can be found in this way:\[x=\frac{b}{2a}\]3. There is no solution of this equation in case if \(b^24ac<0\). Proof: Divide both parts of the equation by \(a, \ a\ne0\):\[x^2+\frac bax+\frac ca=0\]Next we write \(\frac ba=2\cdot\frac b{2a}\). And then add and subtract \(\frac{b^2}{4a^2}\) in the left part of the equation:\[x^2+2\cdot\frac b{2a}x+\frac{b^2}{4a^2}\frac{b^2}{4a^2}+\frac ca=0\]Now use the formula \((a+b)^2=a^2+2ab+b^2\) to the first 3 addends in the left part of the equation. Sum 2 others \(\frac{b^2}{4a^2}+\frac ca=\frac{b^24ac}{4a^2}\) and place them into the right part of the equation: \[\left( x+\frac b{2a}\right)^2=\frac{b^24ac}{4a^2}\]\(4a^2\) is always positive. So, the quantity of the roots depends only on the sign of the radicand. 1. If \(b^24ac>0\) we can take a square root of the both parts of the equation and then add \(\frac b{2a}\) to the both parts and get: \[x=\sqrt{\frac{b^24ac}{4a^2}}\frac b{2a}=\frac {b\pm\sqrt{b^24ac}}{2a}\]The first solution is for \(\) sign, the second is for \(+\). 2. If \(b^24ac=0\) we have an equation \[\left( x+\frac b{2a}\right)^2=0\]And have only 1 solution: \(x=\frac b{2a}\). 3. If \(b^24ac<0\), we cant take a square root to find \(x\), because we cant find a square root of the negative number. In this case there is no solution.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.9Example: 1. \(x^23x+2=0\) Notice that \(a=1, \ b=3, \ c=2\) Check \[b^24ac=(3)^24\cdot1\cdot2=98=1>0\] So, we have 2 solutions. And use the formula: \[x_1=\frac{(3)\sqrt{(3)^24\cdot1\cdot2}}{2\cdot1}=\frac{3\sqrt{1}}{2}=\frac{31}{2}=\frac{2}{2}=1\]\[x_2=\frac{(3)+\sqrt{(3)^24\cdot1\cdot2}}{2\cdot1}=\frac{3+\sqrt{1}}{2}=\frac{3+1}{2}=\frac{4}{2}=2\] So, the solution is \(x_1=1, \ x_2=2\). 2. \(2x^2+x+1=0\) Notice that \(a=2, \ b=1, \ c=1\) Check \[b^24ac=1^24\cdot2\cdot1=18=7<0\]So, we dont have real solution. 3. \(x^2+10x+25=0\) Notice that \(a=1, \ b=10, \ c=25\) Check \[b^24ac=10^24\cdot1\cdot25=100100=0\]So, we have 1 solution: \[x=\frac b{2a}=\frac{10}{2\cdot1}=5\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/4ff461cfe4b01c7be8c7b1ea http://openstudy.com/study#/updates/4fe09fa0e4b06e92b86fbb47 http://openstudy.com/study#/updates/4fe08815e4b06e92b86fa074

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0these are some .. tutorials made by me of quadratic equation ... . well must say nice one

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0So u proved in this tutorial that a quadratic eqn has 2 roots (max.) right ?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.9I think this follows from the properties of the square root. If the statement: "Equation \(x^2=a\) has no more than 2 roots for any real \(a\)" is proved, your statement will follow from this one. I saw you proof of this statement. It is good.
Ask your own question
Sign UpFind 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.