find-a-common-roots-from-the-two-quadratic-eq-24x-2-p-4-x-1-0-and-6x-2-11x-p-2-0-

Question Number 96189 by bemath last updated on 30/May/20

find a common roots from

the two quadratic eq

{24 x}^{2} + (p + 4) x - 1 = 0

and {6 x}^{2} + 11 x + p + 2 = 0

Answered by john santu last updated on 30/May/20

by {Wildberger}^{'} s theorem

if two quadratic eq : x^{2} + Ax + B = 0

and x^{2} + Cx + D = 0 have a common

roots {it}^{'} s \Rightarrow x = \frac{D - B}{A - C}

(i) x^{2} + (\frac{p + 4}{24}) x - \frac{1}{24} = 0

(ii) x^{2} + (\frac{11}{6}) x + (\frac{p + 2}{6}) = 0

s o t h e c o m m o n r o o t s

i s x = \frac{\frac{p + 2}{6} + \frac{1}{24}}{(\frac{p + 4}{24}) - \frac{11}{6}} = \frac{4 p + 9}{p + 4 - 44}

x = \frac{4 p + 9}{p - 40}

Commented by bemath last updated on 30/May/20

wow ... I just found out the theorem mister

Commented by mr W last updated on 30/May/20

o n l y f o r c e r t a i n v a l u e s o f p h a v e b o t h

e q u a t i o n s a c o m m o n r o o t .

Commented by john santu last updated on 30/May/20

Yes, right

Answered by mr W last updated on 30/May/20

{l e t}^{'} s s a y e q n . 1 h a s r o o t s : α, β

e q n . 2 h a s r o o t s : β, γ

α + β = - \frac{p + 4}{24}

α β = - \frac{1}{24}

β + γ = - \frac{11}{6}

β γ = \frac{p + 2}{6}

\Rightarrow (\frac{p + 4}{24} + β) β = \frac{1}{24} \dots (i)

\Rightarrow (\frac{11}{6} + β) β = - \frac{p + 2}{6} \dots (i i)

f r o m (i i) :

\frac{p + 4}{24} = \frac{1}{12} - \frac{1}{4} (\frac{11}{6} + β) β

p u t t h i s i n t o (i) :

6 β^{3} - 13 β^{2} - 2 β + 1 = 0

\Rightarrow (3 β + 1) (2 β^{2} - 5 β + 1) = 0

\Rightarrow 3 β + 1 = 0 \Rightarrow β = - \frac{1}{3}

\Rightarrow 2 β^{2} - 5 β + 1 = 0 \Rightarrow β = \frac{5 \pm \sqrt{17}}{4}

i . e . t h e c o m m o n r o o t c a n b e

- \frac{1}{3}, \frac{5 - \sqrt{17}}{4}, \frac{5 + \sqrt{17}}{4}

p = - 2 - 6 (\frac{11}{6} + β) β

\Rightarrow p = 1

\Rightarrow p = - \frac{63 - 13 \sqrt{17}}{2} \approx - 4.7

\Rightarrow p = - \frac{63 + 13 \sqrt{17}}{2} \approx - 58.3

Answered by 1549442205 last updated on 30/May/20

Denoting x_{0} common root of two equations . Then

{\begin{cases} {24 x}_{0}^{2} + (p + 4) x_{0} - 1 = 0 (1) \\ {6 x}_{0}^{2} + {11 x}_{0} + p + 2 = 0 (2) \end{cases}

\Leftrightarrow {\begin{cases} {24 x}_{0}^{2} + (p + 4) x_{0} - 1 = 0 \\ {24 x}_{0} + {44 x}_{0} + 4 p + 8 = 0 \end{cases}

\Rightarrow (p + 4) x_{0} - 1 = {44 x}_{0} + 4 p + 8

\Rightarrow (p - 40) x_{0} = 4 p + 9 \Rightarrow x_{0} = \frac{4 p + 9}{p - 40} . Replace into (2)

\Rightarrow 6 {(\frac{4 p + 9}{p - 40})}^{2} + 11 (\frac{4 p + 9}{p - 40}) + p + 2 = 0

\Leftrightarrow 6 ({16 p}^{2} + 72 p + 81) + 11 ({4 p}^{2} - 151 p - 360)

+ (p + 2) (p^{2} - 80 p + 1600) = 0 \Leftrightarrow

{140 p}^{2} - 1229 p - 3474 + p^{3} - {78 p}^{2} + 1440 p

+ 3200 = 0 \Leftrightarrow p^{3} + {62 p}^{2} + 211 p - 274 = 0

\Leftrightarrow (p - 1) (p^{2} + 63 p + 274) = 0

a / p = 1 \Rightarrow x_{0} = \frac{- 1}{3}

b / p^{2} + 63 p + 274 = 0 \Leftrightarrow p = \frac{- 63 \pm 13 \sqrt{17}}{2}

\Rightarrow x_{0} \in {\frac{5 - \sqrt{17}}{4}; \frac{\sqrt{17} + 5}{4}}

Thus, two given equations have three

common