| Version 5 |
Version 4 |
| The roots of the quadratic equation |
The roots of the quadratic equation |
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$$ax^2+bx+c=0\qquad{a,b,c\in\R,a\neq 0}$$
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$$ax^2+bx+c=0$$
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| are given by the following formula: |
are given by the following formula: |
| $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ |
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ |
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| The number $\Delta=b^2-4ac$ is called the \emph{discriminant} if the equation. |
The number $\Delta=b^2-4ac$ is called the \emph{discriminant} if the equation. |
| If $\Delta>0$, there are two different real roots, if $\Delta=0$ there is a single real root (counted twice) and if $\Delta<0$ there are no real roots (but two different complex roots). |
If $\Delta>0$, there are two different real roots, if $\Delta=0$ there is a single real root (counted twice) and if $\Delta<0$ there are no real roots (but two different complex roots). |
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| Let's work a few examples. |
Let's work a few examples. |
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| First, consider $2x^2-14x+24=0$. |
First, consider $2x^2-14x+24=0$. |
| Here $a=2,b=-14,c=24$. Substituting in the formula gives us |
Here $a=2,b=-14,c=24$. Substituting in the formula gives us |
| $$x=\frac{14\pm \sqrt{(-14)^2-2\cdot4\cdot24}}{2\cdot 2}=\frac{14\pm\sqrt{4}}{4}=\frac{14\pm2}{4}$$ |
$$x=\frac{14\pm \sqrt{(-14)^2-2\cdot4\cdot24}}{2\cdot 2}=\frac{14\pm\sqrt{4}}{4}=\frac{14\pm2}{4}$$ |
| So we have two solutions (depending if you take the sign $+$ or $-$): |
So we have two solutions (depending if you take the sign $+$ or $-$): |
| $x=\frac{16}{4}=4$ and $x=\frac{12}{4}=3$. |
$x=\frac{16}{4}=4$ and $x=\frac{12}{4}=3$. |
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| Now we will solve $x^2-x-1=0$. |
Now we will solve $x^2-x-1=0$. |
| Here $a=1,b=-1,c=-1$ so |
Here $a=1,b=-1,c=-1$ so |
| $$x=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2}=\frac{1\pm{\sqrt{5}}}{2}$$ |
$$x=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2}=\frac{1\pm{\sqrt{5}}}{2}$$ |
| so the solutions are $x=\frac{1+\sqrt{5}}{2}$ and $x=\frac{1-\sqrt{5}}{2}$. |
so the solutions are $x=\frac{1+\sqrt{5}}{2}$ and $x=\frac{1-\sqrt{5}}{2}$. |
