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Question Number 28319 by NECx last updated on 24/Jan/18

I f t h e r o o t s o f x^{2} + p x + q = 0, q \neq 0

a r e α a n d β . F i n d t h e r o o t s o f

{q x}^{2} + (2 q - p^{2}) x + q = 0 i n t e r m s o f

α a n d β .

Answered by Rasheed.Sindhi last updated on 24/Jan/18

x^{2} + p x + q = 0 \Rightarrow α + β = - p, α β = q

\Rightarrow p = - (α + β), q = α β

{q x}^{2} + (2 q - p^{2}) x + q = 0

x = \frac{- (2 q - p^{2}) \pm \sqrt{{(2 q - p^{2})}^{2} - 4 q^{2}}}{2 q}

x = \frac{- (2 q - p^{2}) \pm \sqrt{4 q^{2} - 4 p^{2} q + p^{4} - 4 q^{2}}}{2 q}

x = \frac{- (2 q - p^{2}) \pm \sqrt{- 4 p^{2} q + p^{4}}}{2 q}

x = \frac{- 2 q + p^{2} \pm p \sqrt{p^{2} - 4 q}}{2 q}

x = \frac{- 2 (α β) + {(α + β)}^{2} \pm (- α - β) \sqrt{{(α + β)}^{2} - 4 α β}}{2 α β}

x = \frac{- 2 α β + {(α + β)}^{2} \mp (α + β) \sqrt{(α^{2} - 2 α β + β^{2}}}{2 α β}

x = \frac{- 2 α β + α^{2} + 2 α β + β^{2} \mp (α + β) (α - β)}{2 α β}

x = \frac{α^{2} + β^{2} \mp (α^{2} - β^{2})}{2 α β}

x = \frac{α^{2} + β^{2} - (α^{2} - β^{2})}{2 α β}, \frac{α^{2} + β^{2} + (α^{2} - β^{2})}{2 α β}

x = \frac{α^{2} + β^{2} - α^{2} + β^{2})}{2 α β}, \frac{α^{2} + β^{2} + α^{2} - β^{2}}{2 α β}

x = \frac{2 β^{2}}{2 α β}, \frac{2 α^{2}}{2 α β}

x = \frac{β}{α}, \frac{α}{β}

Answered by Rasheed.Sindhi last updated on 24/Jan/18

x^{2} + p x + q = 0

p = - (α + β), q = α β

{q x}^{2} + (2 q - p^{2}) x + q = 0 \Rightarrow

α β x^{2} + {2 α β - {(α + β)}^{2}} x + α β = 0

Let the roots are A & B

A + B = - \frac{2 α β - {(α + β)}^{2}}{α β} = \frac{α^{2} + β^{2}}{α β}

AB = \frac{α β}{α β} = 1 \Rightarrow B = \frac{1}{A}

A + B = A + \frac{1}{A} = \frac{α^{2} + β^{2}}{α β} = \frac{α}{β} + \frac{β}{α}

A^{2} - (\frac{α}{β} + \frac{β}{α}) A + 1 = 0

A^{2} - \frac{α}{β} A - \frac{β}{α} A + (\frac{α}{β}) (\frac{β}{α}) = 0

A (A - \frac{α}{β}) - \frac{β}{α} (A - \frac{α}{β}) = 0

(A - \frac{α}{β}) (A - \frac{β}{α}) = 0

A = \frac{α}{β}, \frac{β}{α}

B = \frac{1}{A} = \frac{β}{α}, \frac{α}{β}

Roots are \frac{α}{β} & \frac{β}{α}