If the equation x^2 −cx+d=0 has roots equal to the fourth powers of the roots of x^2 +ax+b=0, where a^2 >4b then the roots of x^2 −4bx+2b^2 −c=0 will be


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Question Number 98058 by PengagumRahasiamu last updated on 11/Jun/20

$If the equation x^{2} - c x + d = 0 has roots$ $equal to the fourth powers of the roots$ $of x^{2} + a x + b = 0, where a^{2} > 4 b then the$ $roots of x^{2} - 4 b x + 2 b^{2} - c = 0 will be$
	Answered by Rio Michael last updated on 11/Jun/20

	$Both real (one is positive and the other negative)$ $see how :$ $let x^{2} + a x + b = 0 have roots α and β$ $then the roots of x^{2} - c x + d = 0 are α^{4} and β^{4}$ $α + β = - a and α β = b$ $also α^{4} + β^{4} = c and α^{4} β^{4} = d$ $\Rightarrow b^{4} = d and α^{4} + β^{4} = c$ ${(α^{2} + β^{2})}^{2} - 2 {(α β)}^{2} = c$ $[{(α + β)}^{2} - 2 α β] - 2 {(α β)}^{2} = c$ $(a^{2} - 2 b^{2}) - 2 b^{2} = c \Rightarrow 2 b^{2} + c = {(a^{2} - 2 b)}^{2}$ $2 b^{2} - c = 4 a^{2} b - a^{2}$ $= a^{2} (4 b - a^{2})$ $now for x^{2} - 4 b x + 2 b^{2} - c = 0$ $discriminant D = {(4 b)}^{2} - 4 (1) (2 b^{2} - c)$ $= 16 b^{2} - 8 b^{2} + 4 c$ $= 8 b^{2} + 4 c$ $= 4 (2 b^{2} + c)$ $= 4 {(a^{2} - 2 b)}^{2} > 0 \Rightarrow real roots$ $f (0) = 2 b^{2} - c = a^{2} (4 b - a^{2}) < 0 \Rightarrow roots have opposite sign$
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