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

⋰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

$$\:\:\:\:\:\iddots\mathcal{B}{e}\mathcal{M}{ath}\iddots \\ $$$${Given}\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}{px}+\mathrm{14}{p}−\mathrm{2}=\mathrm{0} \\ $$$${has}\:{the}\:{roots}\:{are}\:\:{u}\:\&\:{v}\:{where}\:{u},{v}\:\notin\mathbb{Z} \\ $$$${If}\:{u},{v}\:\geqslant\:\mathrm{1}\:,\:{then}\:{the}\:{value}\:{of}\:\mid{u}−{v}\mid\:. \\ $$$$\left({a}\right)\mathrm{14}\:\:\:\:\:\left({b}\right)\mathrm{15}\:\:\:\:\:\left({c}\right)\mathrm{16}\:\:\:\:\:\left({d}\right)\mathrm{17}\:\:\:\left({e}\right)\:\mathrm{18} \\ $$

Commented by Her_Majesty last updated on 10/Aug/20

u=(p/2)−((√(9p^2 −84p+12))/6) v=(p/2)+((√(9p^2 −84p+12))/6) u≥1∧v≥1 ⇒ p≥((14+2(√(46)))/3) ∣u−v∣=((√(9p^2 −84p+12))/3) and we can find a p for any given value of A with A=((√(9p^2 −84p+12))/3) ⇔ p=((14+(√(9A^2 +184)))/3) and p≥((14+2(√(46)))/3) so either the question is strange or I am wrong (where?)

$${u}=\frac{{p}}{\mathrm{2}}−\frac{\sqrt{\mathrm{9}{p}^{\mathrm{2}} −\mathrm{84}{p}+\mathrm{12}}}{\mathrm{6}} \\ $$$${v}=\frac{{p}}{\mathrm{2}}+\frac{\sqrt{\mathrm{9}{p}^{\mathrm{2}} −\mathrm{84}{p}+\mathrm{12}}}{\mathrm{6}} \\ $$$${u}\geqslant\mathrm{1}\wedge{v}\geqslant\mathrm{1}\:\Rightarrow\:{p}\geqslant\frac{\mathrm{14}+\mathrm{2}\sqrt{\mathrm{46}}}{\mathrm{3}} \\ $$$$\mid{u}−{v}\mid=\frac{\sqrt{\mathrm{9}{p}^{\mathrm{2}} −\mathrm{84}{p}+\mathrm{12}}}{\mathrm{3}} \\ $$$${and}\:{we}\:{can}\:{find}\:{a}\:{p}\:{for}\:{any}\:{given}\:{value}\:{of}\:{A} \\ $$$${with} \\ $$$${A}=\frac{\sqrt{\mathrm{9}{p}^{\mathrm{2}} −\mathrm{84}{p}+\mathrm{12}}}{\mathrm{3}}\:\Leftrightarrow\:{p}=\frac{\mathrm{14}+\sqrt{\mathrm{9}{A}^{\mathrm{2}} +\mathrm{184}}}{\mathrm{3}} \\ $$$${and}\:{p}\geqslant\frac{\mathrm{14}+\mathrm{2}\sqrt{\mathrm{46}}}{\mathrm{3}} \\ $$$${so}\:{either}\:{the}\:{question}\:{is}\:{strange}\:{or}\:{I}\:{am} \\ $$$${wrong}\:\left({where}?\right) \\ $$

Commented by bemath last updated on 10/Aug/20

Commented by bemath last updated on 10/Aug/20

$${the}\:{original}\:{question} \\ $$

Commented by bemath last updated on 10/Aug/20

if u+v ≥ 2 ⇒ p ≥ 2 u×v ≥ 1 ⇒((14p−2)/6) ≥ 1 ⇒p ≥ (4/7) (1)∧(2) ⇒ p ≥ 2

$${if}\:{u}+{v}\:\geqslant\:\mathrm{2}\:\Rightarrow\:{p}\:\geqslant\:\mathrm{2} \\ $$$${u}×{v}\:\geqslant\:\mathrm{1}\:\Rightarrow\frac{\mathrm{14}{p}−\mathrm{2}}{\mathrm{6}}\:\geqslant\:\mathrm{1}\:\Rightarrow{p}\:\geqslant\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\left(\mathrm{1}\right)\wedge\left(\mathrm{2}\right)\:\Rightarrow\:{p}\:\geqslant\:\mathrm{2} \\ $$

Commented by bemath last updated on 10/Aug/20