BeMath-Given-6x-2-6px-14p-2-0-has-the-roots-are-u-amp-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 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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