May 23rd, 2018, 08:51 AM  #11  
Senior Member Joined: May 2016 From: USA Posts: 1,038 Thanks: 423  Quote:
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: 19,058 Thanks: 1618 
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. 

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