+2441 BOSEOK is correct; the equation is x=3.
Contrary to what Gh0sty stated, the slope is defined when a linear function happens to be horizontal. Let's figure out the slope of the graph provided by Gh0sty. I will use the points \((3,-2)\) and \((0,-2)\). Let's find the slope.
Of course, the formula for slope is the following:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
| \(m=\frac{-2-(-2)}{0-3}\) | Continue to simplify the fraction. |
| \(m=\frac{0}{-3}=0\) | |
The slope is 0. 0 is a valid number for the slope. If you type into Demos y=0x-2, the line will appear exactly as the picture above.
Let's try calculating the slope of the line x=-3. Two arbitrary points on the line are \((-3,2)\) and \((-3,0)\)
What is the slope of this? Let's try using the formula again. Let's see what happens.
| \(\frac{0-2}{-3-(-3)}\) | Simplify both the numerator and the denominator. |
| \(\frac{-2}{0}\) | |
Of course, any number divided by 0 is undefined and therefore it has an undefined slope.
Therefore, BOSEOK's answer is correct because it meets both criteria of
1) A line with an undefined slope
2) A line that passes through the point \((3,-2)\)
And therefore, \(x=3\) is correct.
+2441 I'll show my method of proving. Of course, the standard form of a quadratic equation is in the following form of \(ax^2+bx+c=0\). If I start with the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), then all I should have to do is utilize some algebraic manipulation to arrive to the standard form. Let's do that!
| \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) | First, let's eliminate the fraction on the right hand side by multiplying by 2a on both sides. |
| \(2ax=-b\pm\sqrt{b^2-4ac}\) | Add b to both sides of the equation. |
| \(2ax+b=\pm\sqrt{b^2-4ac}\) | Square both sides. Normally, you would have to consider both cases, but squaring always results in the positive answer--no matter if the original number is positive or negative. Therefore, both will result to the same thing. |
| \((2ax+b)^2=b^2-4ac\) | Expand the left hand side of the equation by knowing that \((x+y)^2=x^2+2xy+y^2\). |
| \((2ax)^2+2(2ax)(b)+b^2=b^2-4ac\) | Subtract b^2 from both sides. |
| \((2ax)^2+2(2ax)(b)=-4ac\) | Simplify the left hand side by dealing with the multiplication and the exponent. |
| \(4a^2x^2+4axb=-4ac\) | Divide by the GCF of both sides, which is 4a. |
| \(ax^2+bx=-c\) | And finally, subtract c on both sides. |
| \(ax^2+bx+c=0\) | |
I have now proven that the quadratic formula works for all numbers for A,B, and C.
