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May 23rd, 2018, 08:51 AM   #11
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Quote:
 Originally Posted by skipjack When calculating the roots using the standard formula, first choose a sign to replace "±" so that both terms in the numerator have the same sign. If the numerator then has value f, the roots are f/(2a) and 2c/f. Doing it this way avoids excessive rounding error in certain cases.
$ax^2 + bx + c \text { and } a \ne 1 \implies d = b^2 - 4ac \text { and }x = \dfrac{-\ b \pm \sqrt{d}}{2a}.$

Standard form.

Conditions: $c \ne 0 \text { if } b= 0.$

$f = -\ b + \sqrt{d} \text { if } b < 0; \text { else } f = -\ b - \sqrt{d}.$

$\text {Case I: } b < 0.$

$\dfrac{f}{2a} = \dfrac{-\ b + \sqrt{d}}{2a}.$

$\dfrac{2c}{2f} = \dfrac{2c}{-\ b + \sqrt{d}} = \dfrac{2c}{-\ b + \sqrt{d}} * \dfrac{-\ b - \sqrt{d}}{-\ b - \sqrt{d}} =$

$\dfrac{2c(-\ b - \sqrt{d})}{b^2 - d} = \dfrac{2c(-\ b - \sqrt{d})}{b^2 - (b^2 - 4ac)} = \dfrac{2c(-\ b - \sqrt{d})}{4ac} = \dfrac{-\ b - \sqrt{d}}{2a}.$

$\text {Case II: } b \ge 0.$

$\dfrac{f}{2a} = \dfrac{-\ b - \sqrt{d}}{2a}.$

$\dfrac{2c}{2f} = \dfrac{2c}{-\ b - \sqrt{d}} = \dfrac{2c}{-\ b - \sqrt{d}} * \dfrac{-\ b + \sqrt{d}}{-\ b + \sqrt{d}} =$

$\dfrac{2c(-\ b + \sqrt{d})}{b^2 - d} = \dfrac{2c(-\ b + \sqrt{d})}{b^2 - (b^2 - 4ac)} = \dfrac{2c(-\ b + \sqrt{d})}{4ac} = \dfrac{-\ b + \sqrt{d}}{2a}.$

Neat.

So I am guessing that this is useful when $b^2 \approx 4ac.$

Is that correct?

 May 23rd, 2018, 10:12 PM #12 Global Moderator   Joined: Dec 2006 Posts: 19,295 Thanks: 1686 I had in mind when b² is much larger than 4ac. For example, the equation x² - 9876x + 1 = 0 has discriminant d = 97535372. Calculating using intermediate results rounded to 8 significant figures, the larger root is 9876/2 + √(97535372)/2 = 9875.9999. Hence the smaller root is 1/9875.9999 = 0.00010125557, but calculating it as 9876/2 - √(97535372)/2 = 4938 - 4937.9999 gives 0.0001. Thanks from JeffM1

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