⋰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


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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 byHer_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 bybemath last updated on 10/Aug/20

Commented bybemath 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 bybemath 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 bybemath 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 bybemath 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 byhgrocks last updated on 10/Aug/20

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum