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Question Number 70399 by Henri Boucatchou last updated on 04/Oct/19

Solve x^4 + x^3 −2ax^2 −ax + a^2 = 0, a ∈ R

$$\boldsymbol{{Solve}}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} =\:\mathrm{0},\:\:\boldsymbol{{a}}\:\in\:\mathbb{R} \\ $$

Commented by Prithwish sen last updated on 04/Oct/19

x^4 −2ax^2 +a^2 +x^3 −ax =0 (x^2 −a)^2 +x(x^2 −a) =0 (x^2 −a)(x^2 +x−a)=0 ⇒x=±(√)a and x = ((−1±(√(1+4a)))/2) I have not checked the answer.Please check it.

$$\mathrm{x}^{\mathrm{4}} −\mathrm{2ax}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} −\mathrm{ax}\:=\mathrm{0} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{a}\right)^{\mathrm{2}} +\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{a}\right)\:=\mathrm{0} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{a}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\pm\sqrt{}\mathrm{a}\:\mathrm{and}\:\mathrm{x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4a}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{checked}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{answer}}.\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{check}}\:\boldsymbol{\mathrm{it}}. \\ $$

Commented by Prithwish sen last updated on 04/Oct/19

Thank you very much Sir. I correct it.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}.\:\mathrm{I}\:\mathrm{correct}\:\mathrm{it}. \\ $$

Commented by MJS last updated on 04/Oct/19

you forgot the “(√( ))”

$$\mathrm{you}\:\mathrm{forgot}\:\mathrm{the}\:``\sqrt{\:\:}'' \\ $$

Answered by MJS last updated on 04/Oct/19

trying factors of a^2 x∈±{a^2 , a, (√a)} ⇒ x=±(√a) ⇒ x^4 +x^3 −2ax^2 −ax+a^2 = =(x−(√a))(x+(√a))(x^2 +x−a)= =(x−(√a))(x+(√a))(x+(1/2)−((√(4a+1))/2))(x+(1/2)+((√(4a+1))/2)) ⇒ x=±(√a)∨x=−(1/2)±((√(4a+1))/2) ⇒ { ((4 real solutions for 0≤a)),((2 real and 2 imaginary solutions for −(1/4)≤a<0)),((2 conjugated complex and 2 imaginary solutions for a<−(1/4))) :}

$$\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:{a}^{\mathrm{2}} \\ $$$${x}\in\pm\left\{{a}^{\mathrm{2}} ,\:{a},\:\sqrt{{a}}\right\} \\ $$$$\Rightarrow\:{x}=\pm\sqrt{{a}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{2}{ax}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} = \\ $$$$=\left({x}−\sqrt{{a}}\right)\left({x}+\sqrt{{a}}\right)\left({x}^{\mathrm{2}} +{x}−{a}\right)= \\ $$$$=\left({x}−\sqrt{{a}}\right)\left({x}+\sqrt{{a}}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:{x}=\pm\sqrt{{a}}\vee{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{a}+\mathrm{1}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{4}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{a}}\\{\mathrm{2}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{imaginary}\:\mathrm{solutions}\:\mathrm{for}\:−\frac{\mathrm{1}}{\mathrm{4}}\leqslant{a}<\mathrm{0}}\\{\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{and}\:\mathrm{2}\:\mathrm{imaginary}\:\mathrm{solutions}\:\mathrm{for}\:{a}<−\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$

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