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Question Number 207546 by pete last updated on 18/May/24
The real roots of the equation x^2 +6x+c=0 differ by 2n, where n is a real non−zero. Show that n^2 =9−c Given that the roots also have opposite signs, find the set of possible values of n
$$\mathrm{The}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{differ}\:\mathrm{by}\:\mathrm{2n},\:\mathrm{where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{non}−\mathrm{zero}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{n}^{\mathrm{2}} =\mathrm{9}−\mathrm{c} \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{also}\:\mathrm{have}\:\mathrm{opposite} \\ $$$$\mathrm{signs},\:\mathrm{find}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n} \\ $$
Answered by A5T last updated on 18/May/24
x_1 +x_2 =−6; x_1 x_2 =c; ∣x_1 −x_2 ∣=2n ⇒4n^2 =(x_1 −x_2 )^2 =(x_1 +x_2 )^2 −4x_1 x_2 ⇒4n^2 =36−4c⇒n^2 =9−c x_(1,2) =((−6+_− (√(36−4c)))/2)=−3+_− (√(9−c))=−3+_− ∣n∣ when[−3+∣n∣≥0 and −3−∣n∣≤0 ⇒∣n∣≥3 and ∣n∣≥−3⇒∣n∣≥3⇒n≥3 or n≤−3..(i) when −3+∣n∣≤0 and −3−∣n∣≥0 ⇒∣n∣≤3 and ∣n∣≤−3⇒∅...(ii) ⇒n≥3 or n≤−3
$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\mathrm{6};\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} ={c};\:\mid{x}_{\mathrm{1}} −{x}_{\mathrm{2}} \mid=\mathrm{2}{n} \\ $$$$\Rightarrow\mathrm{4}{n}^{\mathrm{2}} =\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{n}^{\mathrm{2}} =\mathrm{36}−\mathrm{4}{c}\Rightarrow{n}^{\mathrm{2}} =\mathrm{9}−{c} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{6}\underset{−} {+}\sqrt{\mathrm{36}−\mathrm{4}{c}}}{\mathrm{2}}=−\mathrm{3}\underset{−} {+}\sqrt{\mathrm{9}−{c}}=−\mathrm{3}\underset{−} {+}\mid{n}\mid \\ $$$${when}\left[−\mathrm{3}+\mid{n}\mid\geqslant\mathrm{0}\:{and}\:−\mathrm{3}−\mid{n}\mid\leqslant\mathrm{0}\right. \\ $$$$\Rightarrow\mid{n}\mid\geqslant\mathrm{3}\:{and}\:\mid{n}\mid\geqslant−\mathrm{3}\Rightarrow\mid{n}\mid\geqslant\mathrm{3}\Rightarrow{n}\geqslant\mathrm{3}\:{or}\:{n}\leqslant−\mathrm{3}..\left({i}\right) \\ $$$${when}\:−\mathrm{3}+\mid{n}\mid\leqslant\mathrm{0}\:{and}\:−\mathrm{3}−\mid{n}\mid\geqslant\mathrm{0} \\ $$$$\Rightarrow\mid{n}\mid\leqslant\mathrm{3}\:{and}\:\mid{n}\mid\leqslant−\mathrm{3}\Rightarrow\emptyset…\left({ii}\right) \\ $$$$\Rightarrow{n}\geqslant\mathrm{3}\:{or}\:{n}\leqslant−\mathrm{3} \\ $$
Commented by pete last updated on 18/May/24
Thank you sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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