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Question Number 9378 by tawakalitu last updated on 03/Dec/16

Solve: x^3 − 18x − 32 = 0

$$\mathrm{Solve}: \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{18x}\:−\:\mathrm{32}\:=\:\mathrm{0} \\ $$

Answered by mrW last updated on 03/Dec/16

a=1, b=0, c=−18, d=−32 Δ=b^2 −3ac=−3×1×(−18)=54 k=((9abc−2b^3 −27a^2 d)/(2(√(∣Δ∣^3 ))))=((−27×(−32))/(2(√(54^3 ))))=((4(√6))/9)≈1.09 since Δ>0 and ∣k∣>1, there is a single root: x_1 =((√(54))/3)×(^3 (√(((4(√6))/9)+(√((((4(√6))/9))^2 −1))))+^3 (√(((4(√6))/9)−(√((((4(√6))/9))^2 −1))))) =(√6)×(^3 (√((4(√6)+(√(15)))/9))+^3 (√((4(√6)−(√(15)))/9))) ≈4.94662127666289

$$\mathrm{a}=\mathrm{1},\:\mathrm{b}=\mathrm{0},\:\mathrm{c}=−\mathrm{18},\:\mathrm{d}=−\mathrm{32} \\ $$$$\Delta=\mathrm{b}^{\mathrm{2}} −\mathrm{3ac}=−\mathrm{3}×\mathrm{1}×\left(−\mathrm{18}\right)=\mathrm{54} \\ $$$$\mathrm{k}=\frac{\mathrm{9abc}−\mathrm{2b}^{\mathrm{3}} −\mathrm{27a}^{\mathrm{2}} \mathrm{d}}{\mathrm{2}\sqrt{\mid\Delta\mid^{\mathrm{3}} }}=\frac{−\mathrm{27}×\left(−\mathrm{32}\right)}{\mathrm{2}\sqrt{\mathrm{54}^{\mathrm{3}} }}=\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{9}}\approx\mathrm{1}.\mathrm{09} \\ $$$$\mathrm{since}\:\Delta>\mathrm{0}\:\mathrm{and}\:\mid\mathrm{k}\mid>\mathrm{1},\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{single}\:\mathrm{root}: \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\sqrt{\mathrm{54}}}{\mathrm{3}}×\left(\:^{\mathrm{3}} \sqrt{\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{9}}+\sqrt{\left(\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{9}}\right)^{\mathrm{2}} −\mathrm{1}}}+\:^{\mathrm{3}} \sqrt{\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{9}}−\sqrt{\left(\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{9}}\right)^{\mathrm{2}} −\mathrm{1}}}\right) \\ $$$$=\sqrt{\mathrm{6}}×\left(\:^{\mathrm{3}} \sqrt{\frac{\mathrm{4}\sqrt{\mathrm{6}}+\sqrt{\mathrm{15}}}{\mathrm{9}}}+\:^{\mathrm{3}} \sqrt{\frac{\mathrm{4}\sqrt{\mathrm{6}}−\sqrt{\mathrm{15}}}{\mathrm{9}}}\right) \\ $$$$\approx\mathrm{4}.\mathrm{94662127666289} \\ $$

Commented by tawakalitu last updated on 03/Dec/16

i really appreciate sir.

$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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