## alternate form of the quadratic formula

I remember there was one way where if b is divisible by 2, you can use a simpler formula but I forgot what it was...
like when ax2+bx+c=0; b is divisible by 2

I might be wrong though : )
 Also, to get the alternate form, divide ax2+bx+c = 0 by x2 to get a + b(1/x) + c(1/x)2=0. This is just a quadratic in 1/x with a and c switched around. So, you can switch a and c in the standard quadratic formula to solve for (1/x) and then switch numerator and denominator to get the alternate form for x.

Mentor
 Quote by Imparcticle why doesn't the computer run them the same?
Computers typically work in base 2. For the point of illustration, assume you are working with a computer that uses base 10 for floating point arithmetic and assume it only has 5 significant digits. For example, computing 1.0/3.0 and 200.0/3.0 on this computer yields 0.33333 and 66.667, respectively. Now suppose you need to find the solutions to

$$x^2-315x+1=0$$

You apply the simple version of the quadratic formula:

$$x=\frac{315\pm\sqrt{315^2-4}}2 = \frac{315\pm\sqrt{99221}}2$$

On this computer, $\surd 99221 = 314.99[/tex] (the correct answer: [itex]314.993651\cdots$. The two solutions as calculated on this computer are

\aligned x_1 &= \frac{315-314.99}2 = 0.005 \\ x_2 &= \frac{315+314.99}2 = 315.00 \aligned

The correct values are $0.00317463517\cdots$ and $314.996825\cdots$. The larger of the two values is correct to five decimal places. The smaller of the two has zero significant digits. Now look what happens when you calculate the smaller value using the alternate form:

$$x_1 = \frac 2{315+314.99} = 0.0031747$$

which is correct to four significant digits.
 Why does the accuracy become better when we move the arithmethics to the divisor ?