If α and β are the roots of ax^2 + bx + c = 0. If ((1 − α)/α) and ((1 − β)/β) are the roots of a_1 x^2 + b_1 x + c_1 = 0. If (b_1 /a


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Question Number 202120 by MATHEMATICSAM last updated on 21/Dec/23

$If α and β are the roots of$ ${a x}^{2} + b x + c = 0 . If \frac{1 - α}{α} and \frac{1 - β}{β} are$ $the roots of a_{1} x^{2} + b_{1} x + c_{1} = 0 . If$ $\frac{b_{1}}{a_{1}} = k + \frac{b}{c} then k = ?$
	Answered by cortano12 last updated on 21/Dec/23
	${ ((ax^2 +bx+c=0⇒α+β=−(b/a))),((a_1 x^2 +b_1 x+c_1 =0⇒(1/α)+(1/β)−2=−(b_1 /a_1 ))) :} ⇒ −(b/c) −2 = −(b_1 /a_1 ) ⇒(b/c) + 2 = (b_1 /a_1 ) = k+(b/c) ∴ k = 2$
	${\begin{cases} {a x}^{2} + b x + c = 0 \Rightarrow α + β = - \frac{b}{a} \\ a_{1} x^{2} + b_{1} x + c_{1} = 0 \Rightarrow \frac{1}{α} + \frac{1}{β} - 2 = - \frac{b_{1}}{a_{1}} \end{cases}$ $\Rightarrow - \frac{b}{c} - 2 = - \frac{b_{1}}{a_{1}}$ $\Rightarrow \frac{b}{c} + 2 = \frac{b_{1}}{a_{1}} = k + \frac{b}{c}$ $∴ k = 2$
	Answered by Rasheed.Sindhi last updated on 21/Dec/23

	$\frac{b_{1}}{a_{1}} = k + \frac{b}{c}$ $k = \frac{b_{1}}{a_{1}} - \frac{b}{c}$ $= - (- \frac{b_{1}}{a_{1}}) + (- \frac{b}{a} \div \frac{c}{a})$ $= - (\frac{1 - α}{α} + \frac{1 - β}{β}) + (α + β) \div α β$ $= - (\frac{β - α β + α - α β}{α β}) + \frac{α + β}{α β}$ $= \frac{- (α + β) + 2 α β + (α + β)}{α β} = 2$
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