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

$$\mathrm{Three}\mathrm{point}\mathrm{are}\mathrm{drawn}\mathrm{on}\mathrm{a}\mathrm{straight}$$$$\mathrm{number}\mathrm{line}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}.$$$$\mathrm{Consider}\mathrm{a}\mathrm{quadractic}\mathrm{equation}$$$$x^2+ax+b=0$$$$a=\mathrm{Length}\mathrm{of}\mathrm{line}\mathrm{segment}\mathrm{AB}$$$$b=\mathrm{Length}\mathrm{of}\mathrm{line}\mathrm{segment}\mathrm{BC}$$$$\mathrm{Give}\mathrm{construction}\mathrm{steps}\mathrm{to}\mathrm{identify}\mathrm{a}\mathrm{points}$$$$\mathrm{in}\mathrm{the}\mathrm{plane}\mathrm{for}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}$$$$\mathrm{using}\mathrm{only}\mathrm{ruler}\left(\mathrm{without}\mathrm{any}\mathrm{markings}\right)\mathrm{and}$$$$\mathrm{compass}.$$$$\mathrm{For}\mathrm{real}\mathrm{roots}\mathrm{point}\mathrm{correspoding}\mathrm{to}\mathrm{root}$$$$\mathrm{will}\mathrm{be}\mathrm{on}\mathrm{number}\mathrm{line},\mathrm{for}\mathrm{complex}\mathrm{root}$$$$\mathrm{in}\mathrm{the}\mathrm{plane}\mathrm{where}y\mathrm{coordinate}{\mathrm{\perp }}^r\mathrm{to}\mathrm{number}$$$$\mathrm{line}\mathrm{should}\mathrm{give}\mathrm{the}\mathrm{imaginary}\mathrm{component}.$$$$\mathrm{Coordinate}\mathrm{of}\mathrm{A}\mathrm{are}(0,0).$$$$\mathrm{Assume}\mathrm{circle}\mathrm{of}\mathrm{radius}1\mathrm{can}\mathrm{be}\mathrm{drawn}\mathrm{using}$$$$\mathrm{compass}\mathrm{this}\mathrm{allows}\mathrm{for}\mathrm{lines}\mathrm{of}\mathrm{unit}\mathrm{length}.$$

Answered by Rasheed Soomro last updated on 19/Dec/15

$$\mid \underset{\stackrel{}{-}}{\overline{\mathcal{A}\mathcal{TRY}}}\mid$$$$x=\frac{-a\pm \sqrt{a^2-4b}}{2}$$$$Forrealroots:$$$$a,baregiven.alinesegmentoflength{a}^{2}canbe$$$$acheivedinlightofyouranswerofQ3607.$$$$subtractingfourtimesb.Squarerootisachieveable.$$$$subtractinga..halvethelinesegment\dots$$$$I,llwriteansweindetailatpresentonlyfor$$$$realx\dots .$$$$Assuminga,b>0[a,bareonrightsidesoforigin.$$$$A=(0,0)\left(given\right)$$$$CoordinatesofBandCalong\overrightarrow{ABC}(x-axis)willbe$$$$(a,0)and(a+b,0)respectively.(Assuming$$$$BisbetweenAandC)$$$$⋮$$$$Forcomplexroots:$$$$x^2+ax+b=0$$$$Letx=x_1+i{x}_2$$$${(x_1+i{x}_2)}^2+a(x_1+i{x}_2)+b=0$$$$x_1^2-x_2^2+2x_1{x}_{2}i+{ax}_1+{ax}_{2}i+b=0;a,b\in {\mathbb{R}}^{+}$$$$x_1^2-x_2^2+{ax}_1+b=0\wedge 2x_1{x}_2+{ax}_2=0$$$$2x_1{x}_2+{ax}_2=0\Rightarrow x_2(2x_1+a)=0\Rightarrow x_2=0\vee x_1=-\frac{a}{2}$$$$For{x}_2=0$$$$x_1^2-x_2^2+{ax}_1+b=0\Rightarrow x_1^2+{ax}_1+b=0$$$$x_1=\frac{-a\pm \sqrt{a^2-4b}}{2}$$$$x_1+x_{2}i=\frac{-a\pm \sqrt{a^2-4b}}{2}+0i=\frac{-a\pm \sqrt{a^2-4b}}{2}$$$$\mathbf{points}\mathbf{of}\mathbf{the}\mathbf{roots}\mathbf{are}(\frac{-a\pm \sqrt{a^2-4b}}{2},0)$$$$\mathbf{we}\mathbf{need}\mathbf{the}\mathbf{lengths}\frac{-a+\sqrt{a^2-4b}}{2}\mathbf{and}\frac{-a-\sqrt{a^2-4b}}{2}$$$$\mathbf{which}\mathbf{are}\mathbf{clearly}\mathbf{constructible}.$$$$For{x}_1=-\frac{a}{2}$$$${(-\frac{a}{2})}^2-x_2^2+a(-\frac{a}{2})+b=0$$$$\frac{a^2}{4}-x_2^2-\frac{a^2}{2}+b=0\Rightarrow x_2^2=\frac{a^2}{4}-\frac{a^2}{2}+b$$$$x_2=\pm \sqrt{\frac{a^2-2a^2+4b}{4}}$$$$x_2=\pm \frac{\sqrt{-a^2+4b}}{2}$$$$x_1+x_{2}i=-\frac{a}{2}\pm \left(\frac{\sqrt{-a^2+4b}}{2}\right)i$$$$\mathbf{Points}\mathbf{are}(-\frac{a}{2},\frac{\sqrt{-a^2+4b}}{2})\mathbf{and}(-\frac{a}{2},-\frac{\sqrt{-a^2+4b}}{2})$$$$\mathbf{Lengths}\mathbf{needed}\mathbf{are}-\frac{a}{2}\mathbf{and}\frac{\sqrt{-a^2+4b}}{2}$$$$\mathbf{which}\mathbf{are}\mathbf{constructible}.$$$$$$

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