ab=c let (a−p)(b−q)=0 ⇒ c−(aq+bp)+pq=0 q=((bp−c)/(p−a)) say 4×2=8 (4−3)(2−((6−8)/(−1)))=1×0=0 And if q=b p=((ab−c)/(2b))=0 And if p+q=c (p−a)(c−p)=bp−c p^2 −(a+c−b)p+c(1−a)=0 2p=(a+c−b)±(√((a+c−b)^2 −4c(1−a))) _______________________


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Question Number 140908 by ajfour last updated on 14/May/21

$a b = c$ $l e t (a - p) (b - q) = 0$ $\Rightarrow c - (a q + b p) + p q = 0$ $q = \frac{b p - c}{p - a}$ $s a y 4 \times 2 = 8$ $(4 - 3) (2 - \frac{6 - 8}{- 1}) = 1 \times 0 = 0$ $A n d i f q = b$ $p = \frac{a b - c}{2 b} = 0$ $A n d i f p + q = c$ $(p - a) (c - p) = b p - c$ $p^{2} - (a + c - b) p + c (1 - a) = 0$ $2 p = (a + c - b) \pm \sqrt{{(a + c - b)}^{2} - 4 c (1 - a)}$ $________________________$
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