find a common roots from the two quadratic eq 24x^2 +(p+4)x−1=0 and 6x^2 +11x+p+2=0


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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

	${l e t}^{'} s s a y e q n . 1 h a s r o o t s : α, β$ $e q n . 2 h a s r o o t s : β, γ$ $α + β = - \frac{p + 4}{24}$ $α β = - \frac{1}{24}$ $β + γ = - \frac{11}{6}$ $β γ = \frac{p + 2}{6}$ $\Rightarrow (\frac{p + 4}{24} + β) β = \frac{1}{24} . . . (i)$ $\Rightarrow (\frac{11}{6} + β) β = - \frac{p + 2}{6} . . . (i i)$ $f r o m (i i) :$ $\frac{p + 4}{24} = \frac{1}{12} - \frac{1}{4} (\frac{11}{6} + β) β$ $p u t t h i s i n t o (i) :$ $6 β^{3} - 13 β^{2} - 2 β + 1 = 0$ $\Rightarrow (3 β + 1) (2 β^{2} - 5 β + 1) = 0$ $\Rightarrow 3 β + 1 = 0 \Rightarrow β = - \frac{1}{3}$ $\Rightarrow 2 β^{2} - 5 β + 1 = 0 \Rightarrow β = \frac{5 \pm \sqrt{17}}{4}$ $i . e . t h e c o m m o n r o o t c a n b e$ $- \frac{1}{3}, \frac{5 - \sqrt{17}}{4}, \frac{5 + \sqrt{17}}{4}$ $p = - 2 - 6 (\frac{11}{6} + β) β$ $\Rightarrow p = 1$ $\Rightarrow p = - \frac{63 - 13 \sqrt{17}}{2} \approx - 4.7$ $\Rightarrow p = - \frac{63 + 13 \sqrt{17}}{2} \approx - 58.3$
	Answered by 1549442205 last updated on 30/May/20
	$Denoting x_0 common root of two equations.Then { (( 24x_0 ^2 +(p+4)x_0 −1=0(1))),((6x_0 ^2 +11x_0 +p+2=0 (2))) :} ⇔ { ((24x_0 ^2 +(p+4)x_0 −1=0)),((24x_0 +44x_0 +4p+8=0)) :} ⇒(p+4)x_0 −1=44x_0 +4p+8 ⇒(p−40)x_0 =4p+9⇒x_0 =((4p+9)/(p−40)).Replace into (2) ⇒6(((4p+9)/(p−40)))^2 +11(((4p+9)/(p−40)))+p+2=0 ⇔6(16p^2 +72p+81)+11(4p^2 −151p−360) +(p+2)(p^2 −80p+1600)=0⇔ 140p^2 −1229p−3474+p^3 −78p^2 +1440p +3200=0⇔p^3 +62p^2 +211p−274=0 ⇔(p−1)(p^2 +63p+274)=0 a/p=1⇒x_0 =((−1)/3) b/p^2 +63p+274=0⇔p=((−63±13(√(17)))/2) ⇒x_0 ∈{((5−(√(17)))/4);(((√(17))+5)/4)} Thus,two given equations have three common$
	$Denoting x_{0} common root of two equations . Then$ ${\begin{cases} {24 x}_{0}^{2} + (p + 4) x_{0} - 1 = 0 (1) \\ {6 x}_{0}^{2} + {11 x}_{0} + p + 2 = 0 (2) \end{cases}$ $\Leftrightarrow {\begin{cases} {24 x}_{0}^{2} + (p + 4) x_{0} - 1 = 0 \\ {24 x}_{0} + {44 x}_{0} + 4 p + 8 = 0 \end{cases}$ $\Rightarrow (p + 4) x_{0} - 1 = {44 x}_{0} + 4 p + 8$ $\Rightarrow (p - 40) x_{0} = 4 p + 9 \Rightarrow x_{0} = \frac{4 p + 9}{p - 40} . Replace into (2)$ $\Rightarrow 6 {(\frac{4 p + 9}{p - 40})}^{2} + 11 (\frac{4 p + 9}{p - 40}) + p + 2 = 0$ $\Leftrightarrow 6 ({16 p}^{2} + 72 p + 81) + 11 ({4 p}^{2} - 151 p - 360)$ $+ (p + 2) (p^{2} - 80 p + 1600) = 0 \Leftrightarrow$ ${140 p}^{2} - 1229 p - 3474 + p^{3} - {78 p}^{2} + 1440 p$ $+ 3200 = 0 \Leftrightarrow p^{3} + {62 p}^{2} + 211 p - 274 = 0$ $\Leftrightarrow (p - 1) (p^{2} + 63 p + 274) = 0$ $a / p = 1 \Rightarrow x_{0} = \frac{- 1}{3}$ $b / p^{2} + 63 p + 274 = 0 \Leftrightarrow p = \frac{- 63 \pm 13 \sqrt{17}}{2}$ $\Rightarrow x_{0} \in {\frac{5 - \sqrt{17}}{4}; \frac{\sqrt{17} + 5}{4}}$ $Thus, two given equations have three$ $common$
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