Solve: q^4 − 40q^2 + q + 384 = 0


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Question Number 42776 by Tawa1 last updated on 02/Sep/18

$Solve : q^{4} - {40 q}^{2} + q + 384 = 0$
	Answered by MJS last updated on 02/Sep/18

	$I found no useable exact solution$ $q_{1} \approx - 4.95762$ $q_{2} \approx - 3.94137$ $q_{3} \approx 4.06764$ $q_{4} \approx 4.83135$
	Commented by Tawa1 last updated on 02/Sep/18

	$God bless you sir . Any workings ? ? ?$
	Commented by MJS last updated on 03/Sep/18
	$I always try to find the number of real zeros first draw the function or just use a calculator to get some values in this case we have 4 real solutions try all factors of the constant 384=2^7 ×3 so we have to try ±{1, 2, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 384} sadly none of these fit we can try to put q^4 −40q^2 +q+384=(q−α−(√β))(q−α+(√β))(q−γ−(√δ))(q−γ+(√δ)) or q^4 −40q^2 +q+384= =(q−α−(√β)−(√γ)−(√δ))(q−α−(√β)+(√γ)+(√δ))(q−α+(√β)−(√γ)+(√δ))(q−α+(√β)+(√γ)−(√δ)) and solve for α, β, γ, δ but in this case both didn′t work, which means the solutions can be found but they′re too complex to work with so we have to approximate (use a calculator or you might get mad)$
	$I always try to find the number of real zeros first$ $draw the function or just use a calculator to$ $get some values$ $in this case we have 4 real solutions$ $try all factors of the constant$ $384 = 2^{7} \times 3 so we have to try$ $\pm {1, 2, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 384}$ $sadly none of these fit$ $we can try to put$ $q^{4} - 40 q^{2} + q + 384 = (q - α - \sqrt{β}) (q - α + \sqrt{β}) (q - γ - \sqrt{δ}) (q - γ + \sqrt{δ})$ $or$ $q^{4} - 40 q^{2} + q + 384 =$ $= (q - α - \sqrt{β} - \sqrt{γ} - \sqrt{δ}) (q - α - \sqrt{β} + \sqrt{γ} + \sqrt{δ}) (q - α + \sqrt{β} - \sqrt{γ} + \sqrt{δ}) (q - α + \sqrt{β} + \sqrt{γ} - \sqrt{δ})$ $and solve for α, β, γ, δ$ $but in this case both {didn}^{'} t work, which means$ $the solutions can be found but {they}^{'} re too$ $complex to work with$ $so we have to approximate (use a calculator or$ $you might get mad)$
	Commented by Tawa1 last updated on 03/Sep/18

	$God bless you sir$
	Commented by Tawa1 last updated on 03/Sep/18

	$Sir, if you still have chance help me to send the breakdown of$ $the reduction of polynomial$
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