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Question Number 77119 by necxxx last updated on 03/Jan/20

s u p p o s e t h e e q u a t i o n s x^{2} + p x + 4 = 0

a n d x^{2} + q x + 3 = 0 h a v e a c o m m o n r o o t,

w r i t e t h i s r o o t i n t e r m s o f t h e o t h e r r o o t .

Answered by jagoll last updated on 03/Jan/20

s u p p o s e i s c o m m o n r o o t i s x_{1}

x_{1}^{2} + {p x}_{1} + 4 = 0 (1)

x_{1}^{2} + {q x}_{1} + 3 = 0 (2)

b y (1) - (2) w e g e t

(p - q) x_{1} = - 1 \Rightarrow x_{1} = \frac{1}{q - p}

n o w w e a p p l i e s {V i e t a}^{'} s r u l e

c o n s i d e r x^{2} + p x + 4 = 0 h a s r o o t

x_{1} a n d x_{2} \Rightarrow x_{1} \times x_{2} = 4

x_{2} = 4 (q - p)

c o n s i d e r x^{2} + q x + 3 = 0

h a s r o o t x_{1} a n d x_{3}

x_{3} = 3 (q - p) .

Answered by MJS last updated on 04/Jan/20

let r the common root

(1) \Rightarrow (x - r) (x - \frac{4}{r}) = 0 \Rightarrow p = - (r + \frac{4}{r})

(2) \Rightarrow (x - r) (x - \frac{3}{r}) = 0 \Rightarrow q = - (r + \frac{3}{r})

common root in terms of the other root

(1) let s the other root \Rightarrow s = \frac{4}{r} \Leftrightarrow r = \frac{4}{s}

(2) let t the other root \Rightarrow t = \frac{3}{r} \Leftrightarrow r = \frac{3}{t}

(\Rightarrow \frac{4}{s} = \frac{3}{t} \Leftrightarrow s = \frac{4}{3} t \Leftrightarrow t = \frac{3}{4} s)

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