## klimenkov 3 years ago Tutorial. How to solve the quadratic equation.

1. klimenkov

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. klimenkov

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: $x=-\frac b{2a}=-\frac{10}{2\cdot1}=-5$

3. mathslover
4. mathslover

these are some .. tutorials made by me of quadratic equation ... . well must say nice one

5. mathslover

So u proved in this tutorial that a quadratic eqn has 2 roots (max.) right ?

6. klimenkov

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.

7. mathslover

thanks ..