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klimenkov Group Title

Tutorial. How to solve the quadratic equation.

  • 2 years ago
  • 2 years ago

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  1. klimenkov Group Title
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    Any 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 \(b^2-4ac>0\) and they can be found in this way:\[x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}, \ x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}\]2. There is 1 solution of this equation in case if \(b^2-4ac=0\) and it can be found in this way:\[x=-\frac{b}{2a}\]3. There is no solution of this equation in case if \(b^2-4ac<0\). 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 \(-\frac{b^2}{4a^2}+\frac ca=-\frac{b^2-4ac}{4a^2}\) and place them into the right part of the equation: \[\left( x+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}\]\(4a^2\) is always positive. So, the quantity of the roots depends only on the sign of the radicand. 1. If \(b^2-4ac>0\) 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: \[x=\sqrt{\frac{b^2-4ac}{4a^2}}-\frac b{2a}=-\frac {b\pm\sqrt{b^2-4ac}}{2a}\]The first solution is for \(-\) sign, the second is for \(+\). 2. If \(b^2-4ac=0\) we have an equation \[\left( x+\frac b{2a}\right)^2=0\]And have only 1 solution: \(x=-\frac b{2a}\). 3. If \(b^2-4ac<0\), 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.

    • 2 years ago
  2. klimenkov Group Title
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    Example: 1. \(x^2-3x+2=0\) Notice that \(a=1, \ b=-3, \ c=2\) Check \[b^2-4ac=(-3)^2-4\cdot1\cdot2=9-8=1>0\] So, we have 2 solutions. And use the formula: \[x_1=\frac{-(-3)-\sqrt{(-3)^2-4\cdot1\cdot2}}{2\cdot1}=\frac{3-\sqrt{1}}{2}=\frac{3-1}{2}=\frac{2}{2}=1\]\[x_2=\frac{-(-3)+\sqrt{(-3)^2-4\cdot1\cdot2}}{2\cdot1}=\frac{3+\sqrt{1}}{2}=\frac{3+1}{2}=\frac{4}{2}=2\] 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 \[b^2-4ac=1^2-4\cdot2\cdot1=1-8=-7<0\]So, we dont have real solution. 3. \(x^2+10x+25=0\) Notice that \(a=1, \ b=10, \ c=25\) Check \[b^2-4ac=10^2-4\cdot1\cdot25=100-100=0\]So, we have 1 solution: \[x=-\frac b{2a}=-\frac{10}{2\cdot1}=-5\]

    • 2 years ago
  3. mathslover Group Title
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    these are some .. tutorials made by me of quadratic equation ... . well must say nice one

    • 2 years ago
  4. mathslover Group Title
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    So u proved in this tutorial that a quadratic eqn has 2 roots (max.) right ?

    • 2 years ago
  5. klimenkov Group Title
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    I 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.

    • 2 years ago
  6. mathslover Group Title
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    thanks ..

    • 2 years ago
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