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klimenkov
 2 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
 2 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
 2 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
 2 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
 2 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
 2 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.
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