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Given-that-17-27-4-6-1-3-and-17-27-4-6-1-3-are-the-roots-of-the-equation-x-2-ax-b-0-Find-the-value-of-ab-

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Question Number 116358 by bemath last updated on 03/Oct/20
Given that ((17−((27)/4)(√6)))^(1/(3 )) and ((17+((27)/4)(√6)))^(1/(3 )) are the roots of the equation x^2 −ax+b = 0. Find the value of ab.
$$\mathrm{Given}\:\mathrm{that}\:\sqrt[{\mathrm{3}\:}]{\mathrm{17}−\frac{\mathrm{27}}{\mathrm{4}}\sqrt{\mathrm{6}}}\:\mathrm{and}\:\sqrt[{\mathrm{3}\:}]{\mathrm{17}+\frac{\mathrm{27}}{\mathrm{4}}\sqrt{\mathrm{6}}} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{b}\:=\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{ab}. \\ $$
Answered by MJS_new last updated on 03/Oct/20
((17±((27(√6))/4)))^(1/3) =2±((√6)/2) (x−2+((√6)/2))(x−2−((√6)/2))=x^2 −4x+(5/2) ⇒ ab=+10
$$\sqrt[{\mathrm{3}}]{\mathrm{17}\pm\frac{\mathrm{27}\sqrt{\mathrm{6}}}{\mathrm{4}}}=\mathrm{2}\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\left({x}−\mathrm{2}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)\left({x}−\mathrm{2}−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)={x}^{\mathrm{2}} −\mathrm{4}{x}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\:{ab}=+\mathrm{10} \\ $$
Commented by bemath last updated on 03/Oct/20
typo sir. the equation x^2 −ax+b=0 then a = 4 & b = (5/2) sir
$$\mathrm{typo}\:\mathrm{sir}.\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} −\mathrm{ax}+\mathrm{b}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{a}\:=\:\mathrm{4}\:\&\:\mathrm{b}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\:\mathrm{sir} \\ $$
Commented by MJS_new last updated on 03/Oct/20
yes you′re right
$$\mathrm{yes}\:\mathrm{you}'\mathrm{re}\:\mathrm{right} \\ $$
Commented by bobhans last updated on 03/Oct/20
i like your method prof. it′s great
$$\mathrm{i}\:\mathrm{like}\:\mathrm{your}\:\mathrm{method}\:\mathrm{prof}.\:\mathrm{it}'\mathrm{s}\:\mathrm{great} \\ $$
Answered by bobhans last updated on 03/Oct/20
Let x_1 = ((17−((27)/4)(√6)))^(1/(3 )) and x_2 =((17+((27)/4)(√6)))^(1/(3 )) Then x_1 .x_2 = (((17−((27)/4)(√6))(17+((27)/4)(√6))))^(1/3) = (((125)/8))^(1/(3 )) = (5/2) x_1 +x_2 = a ⇒a^3 = x_1 ^3 + x_2 ^3 +3x_1 x_2 (x_1 +x_2 ) a^3 = 34+3ab ⇒a^3 = 34 +((15)/2)a ⇒2a^3 −15a−68 = 0 ⇒(a−4)(2a^2 +8a+17) =0 , for a∈R we get a = 4 and thus ab = 4×(5/2)=10
$$\mathrm{Let}\:\mathrm{x}_{\mathrm{1}} \:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{17}−\frac{\mathrm{27}}{\mathrm{4}}\sqrt{\mathrm{6}}}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\sqrt[{\mathrm{3}\:}]{\mathrm{17}+\frac{\mathrm{27}}{\mathrm{4}}\sqrt{\mathrm{6}}} \\ $$$$\mathrm{Then}\:\mathrm{x}_{\mathrm{1}} .\mathrm{x}_{\mathrm{2}} \:=\:\sqrt[{\mathrm{3}}]{\left(\mathrm{17}−\frac{\mathrm{27}}{\mathrm{4}}\sqrt{\mathrm{6}}\right)\left(\mathrm{17}+\frac{\mathrm{27}}{\mathrm{4}}\sqrt{\mathrm{6}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{125}}{\mathrm{8}}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \:=\:{a}\:\Rightarrow{a}^{\mathrm{3}} \:=\:{x}_{\mathrm{1}} ^{\mathrm{3}} \:+\:{x}_{\mathrm{2}} ^{\mathrm{3}} \:+\mathrm{3}{x}_{\mathrm{1}} {x}_{\mathrm{2}} \left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{3}} \:=\:\mathrm{34}+\mathrm{3}{ab}\:\Rightarrow{a}^{\mathrm{3}} \:=\:\mathrm{34}\:+\frac{\mathrm{15}}{\mathrm{2}}{a} \\ $$$$\Rightarrow\mathrm{2}{a}^{\mathrm{3}} −\mathrm{15}{a}−\mathrm{68}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left({a}−\mathrm{4}\right)\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{8}{a}+\mathrm{17}\right)\:=\mathrm{0}\:,\:\mathrm{for}\:{a}\in\mathbb{R} \\ $$$$\mathrm{we}\:\mathrm{get}\:{a}\:=\:\mathrm{4}\:\mathrm{and}\:\mathrm{thus}\:{ab}\:=\:\mathrm{4}×\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{10} \\ $$

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