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Three-point-are-drawn-on-a-straight-number-line-A-B-and-C-Consider-a-quadractic-equation-x-2-ax-b-0-a-Length-of-line-segment-AB-b-Length-of-line-segment-BC-Give-construction-steps-to-identify-a-poin

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Question Number 3588 by prakash jain last updated on 16/Dec/15
Three point are drawn on a straight number line A,B and C. Consider a quadractic equation x^2 +ax+b=0 a=Length of line segment AB b=Length of line segment BC Give construction steps to identify a points in the plane for the roots of the above equation using only ruler (without any markings) and compass. For real roots point correspoding to root will be on number line, for complex root in the plane where y coordinate ⊥^r to number line should give the imaginary component. Coordinate of A are (0,0). Assume circle of radius 1 can be drawn using compass this allows for lines of unit length.
$$\mathrm{Three}\:\mathrm{point}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{number}\:\mathrm{line}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{C}. \\ $$$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{quadractic}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${a}=\mathrm{Length}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{AB} \\ $$$${b}=\mathrm{Length}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{BC} \\ $$$$\mathrm{Give}\:\mathrm{construction}\:\mathrm{steps}\:\mathrm{to}\:\mathrm{identify}\:\mathrm{a}\:\mathrm{points} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{for}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above}\:\mathrm{equation} \\ $$$$\mathrm{using}\:\mathrm{only}\:\mathrm{ruler}\:\left(\mathrm{without}\:\mathrm{any}\:\mathrm{markings}\right)\:\mathrm{and} \\ $$$$\mathrm{compass}. \\ $$$$\mathrm{For}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{point}\:\mathrm{correspoding}\:\mathrm{to}\:\mathrm{root} \\ $$$$\mathrm{will}\:\mathrm{be}\:\mathrm{on}\:\mathrm{number}\:\mathrm{line},\:\mathrm{for}\:\mathrm{complex}\:\mathrm{root} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{where}\:{y}\:\mathrm{coordinate}\:\bot^{{r}} \:\mathrm{to}\:\mathrm{number} \\ $$$$\mathrm{line}\:\mathrm{should}\:\mathrm{give}\:\mathrm{the}\:\mathrm{imaginary}\:\mathrm{component}. \\ $$$$\mathrm{Coordinate}\:\mathrm{of}\:\mathrm{A}\:\mathrm{are}\:\left(\mathrm{0},\mathrm{0}\right). \\ $$$$\mathrm{Assume}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{using} \\ $$$$\mathrm{compass}\:\mathrm{this}\:\mathrm{allows}\:\mathrm{for}\:\mathrm{lines}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{length}. \\ $$
Answered by Rasheed Soomro last updated on 19/Dec/15
∣A TRY _(−^ ) ^(−) ∣ x=((−a±(√(a^2 −4b)))/2) For real roots : a,b are given. a line segment of length a^2 can be acheived in light of your answer of Q 3607. subtracting four times b.Square root is achieveable. subtracting a..halve the line segment... I,ll write answe in detail at present only for real x.... Assuming a,b>0 [a , b are on right sides of origin. A=(0,0) (given) Coordinates of B and C along ABC^(→) (x−axis) will be (a,0) and (a+b,0) respectively.(Assuming B is between A and C) ⋮ For complex roots: x^2 +ax+b=0 Let x=x_1 +i x_2 (x_1 +i x_2 )^2 +a(x_1 +i x_2 )+b=0 x_1 ^2 −x_2 ^2 +2x_1 x_2 i+ax_1 +ax_2 i +b=0 ; a,b∈R^+ x_1 ^2 −x_2 ^2 +ax_1 +b=0 ∧ 2x_1 x_2 +ax_2 =0 2x_1 x_2 +ax_2 =0⇒x_2 (2x_1 +a)=0⇒x_2 =0 ∨ x_1 =−(a/2) For x_2 =0 x_1 ^2 −x_2 ^2 +ax_1 +b=0⇒x_1 ^2 +ax_1 +b=0 x_1 =((−a±(√(a^2 −4b)))/2) x_1 +x_2 i=((−a±(√(a^2 −4b)))/2)+0i=((−a±(√(a^2 −4b)))/2) points of the roots are (((−a±(√(a^2 −4b)))/2) , 0) we need the lengths ((−a+(√(a^2 −4b)))/2) and ((−a−(√(a^2 −4b)))/2) which are clearly constructible. For x_1 =−(a/2) (−(a/2))^2 −x_2 ^2 +a(−(a/2))+b=0 (a^2 /4)−x_2 ^2 −(a^2 /2)+b=0⇒x_2 ^2 =(a^2 /4)−(a^2 /2)+b x_2 =±(√((a^2 −2a^2 +4b)/4)) x_2 =±((√(−a^2 +4b))/2) x_1 +x_2 i=− (a/2)±(((√(−a^2 +4b))/2))i Points are (− (a/2) , ((√(−a^2 +4b))/2)) and (− (a/2) ,− ((√(−a^2 +4b))/2)) Lengths needed are − (a/2) and ((√(−a^2 +4b))/2) which are constructible.
$$\mid\underset{\overset{} {−}} {\overline {\boldsymbol{\mathcal{A}}\:\:\boldsymbol{\mathcal{TRY}}\:}}\mid \\ $$$${x}=\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$${For}\:{real}\:{roots}\:: \\ $$$${a},{b}\:{are}\:{given}.\:{a}\:{line}\:{segment}\:{of}\:{length}\:{a}^{\mathrm{2}} \:{can}\:{be}\: \\ $$$${acheived}\:{in}\:{light}\:{of}\:{your}\:{answer}\:{of}\:{Q}\:\mathrm{3607}. \\ $$$${subtracting}\:{four}\:{times}\:{b}.{Square}\:{root}\:{is}\:{achieveable}. \\ $$$${subtracting}\:\:{a}..{halve}\:{the}\:{line}\:{segment}… \\ $$$${I},{ll}\:{write}\:{answe}\:{in}\:{detail}\:{at}\:{present}\:{only}\:{for} \\ $$$${real}\:{x}…. \\ $$$${Assuming}\:{a},{b}>\mathrm{0}\:\left[{a}\:,\:{b}\:{are}\:{on}\:{right}\:{sides}\:{of}\:{origin}.\right. \\ $$$${A}=\left(\mathrm{0},\mathrm{0}\right)\:\left({given}\right) \\ $$$${Coordinates}\:{of}\:{B}\:{and}\:{C}\:\:\:{along}\:\overset{\rightarrow} {{ABC}}\:\left({x}−{axis}\right)\:\:{will}\:{be} \\ $$$$\left({a},\mathrm{0}\right)\:{and}\:\left({a}+{b},\mathrm{0}\right)\:{respectively}.\left({Assuming}\right. \\ $$$$\left.{B}\:{is}\:{between}\:{A}\:{and}\:\:{C}\right) \\ $$$$\vdots \\ $$$${For}\:{complex}\:{roots}: \\ $$$${x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${Let}\:\:{x}={x}_{\mathrm{1}} +{i}\:{x}_{\mathrm{2}} \\ $$$$\left({x}_{\mathrm{1}} +{i}\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} +{a}\left({x}_{\mathrm{1}} +{i}\:{x}_{\mathrm{2}} \right)+{b}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} −{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} {i}+{ax}_{\mathrm{1}} +{ax}_{\mathrm{2}} {i}\:+{b}=\mathrm{0}\:\:\:;\:\:{a},{b}\in\mathbb{R}^{+} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} −{x}_{\mathrm{2}} ^{\mathrm{2}} +{ax}_{\mathrm{1}} +{b}=\mathrm{0}\:\wedge\:\:\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{ax}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{ax}_{\mathrm{2}} =\mathrm{0}\Rightarrow{x}_{\mathrm{2}} \left(\mathrm{2}{x}_{\mathrm{1}} +{a}\right)=\mathrm{0}\Rightarrow{x}_{\mathrm{2}} =\mathrm{0}\:\:\vee\:\:{x}_{\mathrm{1}} =−\frac{{a}}{\mathrm{2}} \\ $$$${For}\:{x}_{\mathrm{2}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} −{x}_{\mathrm{2}} ^{\mathrm{2}} +{ax}_{\mathrm{1}} +{b}=\mathrm{0}\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{2}} +{ax}_{\mathrm{1}} +{b}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} {i}=\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}+\mathrm{0}{i}=\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{points}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{roots}}\:\boldsymbol{\mathrm{are}}\:\left(\frac{−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}\:,\:\mathrm{0}\right) \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{lengths}}\:\:\frac{−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}\:\boldsymbol{\mathrm{and}}\:\frac{−{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{clearly}}\:\boldsymbol{\mathrm{constructible}}. \\ $$$${For}\:{x}_{\mathrm{1}} =−\frac{{a}}{\mathrm{2}} \\ $$$$\left(−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} −{x}_{\mathrm{2}} ^{\mathrm{2}} +{a}\left(−\frac{{a}}{\mathrm{2}}\right)+{b}=\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−{x}_{\mathrm{2}} ^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+{b}=\mathrm{0}\Rightarrow{x}_{\mathrm{2}} ^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+{b} \\ $$$${x}_{\mathrm{2}} =\pm\sqrt{\frac{{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} +\mathrm{4}{b}}{\mathrm{4}}} \\ $$$${x}_{\mathrm{2}} =\pm\frac{\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{b}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} {i}=−\:\frac{{a}}{\mathrm{2}}\pm\left(\frac{\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{b}}}{\mathrm{2}}\right){i} \\ $$$$\boldsymbol{\mathrm{Points}}\:\boldsymbol{\mathrm{are}}\:\left(−\:\frac{{a}}{\mathrm{2}}\:,\:\frac{\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{b}}}{\mathrm{2}}\right)\:\boldsymbol{\mathrm{and}}\:\left(−\:\frac{{a}}{\mathrm{2}}\:,−\:\frac{\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{b}}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{\mathrm{Lengths}}\:\boldsymbol{\mathrm{needed}}\:\boldsymbol{\mathrm{are}}\:−\:\frac{{a}}{\mathrm{2}}\:\:\:\boldsymbol{\mathrm{and}}\:\frac{\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{b}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{constructible}}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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