If the roots of x^2 +px+q=0, q≠0 are α and β.Find the roots of qx^2 +(2q−p^2 )x+q=0 in terms of α and β.


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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
	$x^2 +px+q=0 p=−(α+β) , q=αβ qx^2 +(2q−p^2 )x+q=0⇒ αβx^2 +{2αβ−(α+β)^2 }x+αβ=0 Let the roots are A & B A+B=−((2αβ−(α+β)^2 )/(αβ))=((α^2 +β^2 )/(αβ)) AB=((αβ)/(αβ))=1⇒B=(1/A) A+B=A+(1/A)=((α^2 +β^2 )/(αβ))=(α/β)+(β/α) A^2 −((α/β)+(β/α))A+1=0 A^2 −(α/β)A−(β/α)A+((α/β))((β/α))=0 A(A−(α/β))−(β/α)(A−(α/β))=0 (A−(α/β))(A−(β/α))=0 A=(α/β) , (β/α) B=(1/A)=(β/α) , (α/β) Roots are (α/β) & (β/α)$
	$x^{2} + p x + q = 0$ $p = - (α + β), q = α β$ ${q x}^{2} + (2 q - p^{2}) x + q = 0 \Rightarrow$ $α β x^{2} + {2 α β - {(α + β)}^{2}} x + α β = 0$ $Let the roots are A & B$ $A + B = - \frac{2 α β - {(α + β)}^{2}}{α β} = \frac{α^{2} + β^{2}}{α β}$ $AB = \frac{α β}{α β} = 1 \Rightarrow B = \frac{1}{A}$ $A + B = A + \frac{1}{A} = \frac{α^{2} + β^{2}}{α β} = \frac{α}{β} + \frac{β}{α}$ $A^{2} - (\frac{α}{β} + \frac{β}{α}) A + 1 = 0$ $A^{2} - \frac{α}{β} A - \frac{β}{α} A + (\frac{α}{β}) (\frac{β}{α}) = 0$ $A (A - \frac{α}{β}) - \frac{β}{α} (A - \frac{α}{β}) = 0$ $(A - \frac{α}{β}) (A - \frac{β}{α}) = 0$ $A = \frac{α}{β}, \frac{β}{α}$ $B = \frac{1}{A} = \frac{β}{α}, \frac{α}{β}$ $Roots are \frac{α}{β} & \frac{β}{α}$
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