⋰BeMath⋰ Given 6x^2 −6px+14p−2=0 has the roots are u & v where u,v ∉Z If u,v ≥ 1 , then the value of ∣u−v∣ . (a)14 (b)15 (c)16 (d)17 (e) 18


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Question Number 107385 by bemath last updated on 10/Aug/20

$\iddots B e M a t h \iddots$ $G i v e n 6 x^{2} - 6 p x + 14 p - 2 = 0$ $h a s t h e r o o t s a r e u & v w h e r e u, v \notin Z$ $I f u, v ⩾ 1, t h e n t h e v a l u e o f ∣ u - v ∣ .$ $(a) 14 (b) 15 (c) 16 (d) 17 (e) 18$
Commented by Her_Majesty last updated on 10/Aug/20

$u = \frac{p}{2} - \frac{\sqrt{9 p^{2} - 84 p + 12}}{6}$ $v = \frac{p}{2} + \frac{\sqrt{9 p^{2} - 84 p + 12}}{6}$ $u ⩾ 1 \land v ⩾ 1 \Rightarrow p ⩾ \frac{14 + 2 \sqrt{46}}{3}$ $∣ u - v ∣= \frac{\sqrt{9 p^{2} - 84 p + 12}}{3}$ $a n d w e c a n f i n d a p f o r a n y g i v e n v a l u e o f A$ $w i t h$ $A = \frac{\sqrt{9 p^{2} - 84 p + 12}}{3} \Leftrightarrow p = \frac{14 + \sqrt{9 A^{2} + 184}}{3}$ $a n d p ⩾ \frac{14 + 2 \sqrt{46}}{3}$ $s o e i t h e r t h e q u e s t i o n i s s t r a n g e o r I a m$ $w r o n g (w h e r e ?)$
Commented by bemath last updated on 10/Aug/20

Commented by bemath last updated on 10/Aug/20

$t h e o r i g i n a l q u e s t i o n$
Commented by bemath last updated on 10/Aug/20

$i f u + v ⩾ 2 \Rightarrow p ⩾ 2$ $u \times v ⩾ 1 \Rightarrow \frac{14 p - 2}{6} ⩾ 1 \Rightarrow p ⩾ \frac{4}{7}$ $(1) \land (2) \Rightarrow p ⩾ 2$
Commented by bemath last updated on 10/Aug/20

$h o w w e g e t t h e v a l u e o f u a n d v ?$
Commented by bemath last updated on 10/Aug/20

$t h e r o o t s a r e n o t i n t e g e r$
Commented by hgrocks last updated on 10/Aug/20

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