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

Solve: q^4 − 40q^2 + q + 384 = 0

$$\mathrm{Solve}:\:\:\:\:\:\mathrm{q}^{\mathrm{4}} \:−\:\mathrm{40q}^{\mathrm{2}} \:+\:\mathrm{q}\:+\:\mathrm{384}\:=\:\mathrm{0} \\ $$

Answered by MJS last updated on 02/Sep/18

I found no useable exact solution q_1 ≈−4.95762 q_2 ≈−3.94137 q_3 ≈4.06764 q_4 ≈4.83135

$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution} \\ $$$${q}_{\mathrm{1}} \approx−\mathrm{4}.\mathrm{95762} \\ $$$${q}_{\mathrm{2}} \approx−\mathrm{3}.\mathrm{94137} \\ $$$${q}_{\mathrm{3}} \approx\mathrm{4}.\mathrm{06764} \\ $$$${q}_{\mathrm{4}} \approx\mathrm{4}.\mathrm{83135} \\ $$

Commented by Tawa1 last updated on 02/Sep/18

God bless you sir. Any workings ???

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Any}\:\mathrm{workings}\:??? \\ $$

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)

$$\mathrm{I}\:\mathrm{always}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{zeros}\:\mathrm{first} \\ $$$$\mathrm{draw}\:\mathrm{the}\:\mathrm{function}\:\mathrm{or}\:\mathrm{just}\:\mathrm{use}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{to} \\ $$$$\mathrm{get}\:\mathrm{some}\:\mathrm{values} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have}\:\mathrm{4}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{try}\:\mathrm{all}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{constant} \\ $$$$\mathrm{384}=\mathrm{2}^{\mathrm{7}} ×\mathrm{3}\:\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{try} \\ $$$$\pm\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{4},\:\mathrm{6},\:\mathrm{8},\:\mathrm{12},\:\mathrm{16},\:\mathrm{24},\:\mathrm{32},\:\mathrm{48},\:\mathrm{64},\:\mathrm{96},\:\mathrm{128},\:\mathrm{192},\:\mathrm{384}\right\} \\ $$$$\mathrm{sadly}\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{fit} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{try}\:\mathrm{to}\:\mathrm{put} \\ $$$${q}^{\mathrm{4}} −\mathrm{40}{q}^{\mathrm{2}} +{q}+\mathrm{384}=\left({q}−\alpha−\sqrt{\beta}\right)\left({q}−\alpha+\sqrt{\beta}\right)\left({q}−\gamma−\sqrt{\delta}\right)\left({q}−\gamma+\sqrt{\delta}\right) \\ $$$$\mathrm{or} \\ $$$${q}^{\mathrm{4}} −\mathrm{40}{q}^{\mathrm{2}} +{q}+\mathrm{384}= \\ $$$$=\left({q}−\alpha−\sqrt{\beta}−\sqrt{\gamma}−\sqrt{\delta}\right)\left({q}−\alpha−\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\delta}\right)\left({q}−\alpha+\sqrt{\beta}−\sqrt{\gamma}+\sqrt{\delta}\right)\left({q}−\alpha+\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\delta}\right) \\ $$$$\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:\alpha,\:\beta,\:\gamma,\:\delta \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{both}\:\mathrm{didn}'\mathrm{t}\:\mathrm{work},\:\mathrm{which}\:\mathrm{means} \\ $$$$\mathrm{the}\:\mathrm{solutions}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{but}\:\mathrm{they}'\mathrm{re}\:\mathrm{too} \\ $$$$\mathrm{complex}\:\mathrm{to}\:\mathrm{work}\:\mathrm{with} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{approximate}\:\left(\mathrm{use}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{or}\right. \\ $$$$\left.\mathrm{you}\:\mathrm{might}\:\mathrm{get}\:\mathrm{mad}\right) \\ $$

Commented by Tawa1 last updated on 03/Sep/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\: \\ $$

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

$$\mathrm{Sir},\:\mathrm{if}\:\mathrm{you}\:\mathrm{still}\:\mathrm{have}\:\mathrm{chance}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{send}\:\mathrm{the}\:\mathrm{breakdown}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{reduction}\:\mathrm{of}\:\mathrm{polynomial}\: \\ $$

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