+2446 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.
+9481 A monic polynomial is a polynomial where the cofficient of the highest order term is 1. (I didn't know this until now, I had to look it up here.)
f(x) is a monic polynomial with a degree of 2. So we can say...
f(x) = 1x2 + bx + c = x2 + bx + c
The problem says f(0) = 4 . So...
f(0) = 02 + b(0) + c
4 = 0 + 0 + c
4 = c
Now that we know c = 4 , we know that f(x) = x2 + bx + 4 .
The problem says f(1) = 10 . So...
f(1) = 12 + b(1) + 4
10 = 1 + b + 4
10 = 5 + b
5 = b
Now we know b = 5 and c = 4 , so f(x) = x2 + 5x + 4 .
I hope this helps ( i took a snapshot from word, cause LaTeX gave me cancer 
).
