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Question Number 202459 by MATHEMATICSAM last updated on 27/Dec/23

$$\mathrm{If}\mathrm{the}\mathrm{difference}\mathrm{of}\mathrm{two}\mathrm{roots}\mathrm{of}$$$$x^2-lx+m=0\mathrm{is}1\mathrm{then}\mathrm{prove}\mathrm{that}$$$$l^2+4m^2={(1+2m)}^2.$$

Answered by aleks041103 last updated on 27/Dec/23

$$x_{1,2}=\frac{1}{2}(l\pm \sqrt{l^2-4m})$$$$\Rightarrow \mid x_1-x_2\mid =\sqrt{l^2-4m}=1$$$$\Rightarrow l^2-4m=1$$$$\Rightarrow l^2=1+4m$$$$\Rightarrow l^2+4m^2=1+4m+4m^2=1+2\left(1\right)\left(2m\right)+{\left(2m\right)}^2$$$$\Rightarrow l^2+4m^2={(1+2m)}^2$$

Answered by Rasheed.Sindhi last updated on 27/Dec/23

$$roots:\alpha ,\alpha -1\left(say\right)$$$$\alpha +(\alpha -1)=l\wedge \alpha (\alpha -1)=m$$$$l=2\alpha -1\wedge m=\alpha (\alpha -1)$$$$\bullet l^2+4m^2={(1+2m)}^2$$$$lhs:{(2\alpha -1)}^2+4{\left(\alpha (\alpha -1)\right)}^2$$$$=4{\alpha }^4-8{\alpha }^3+8{\alpha }^2-4\alpha +1$$$$rhs:{(1+2\alpha (\alpha -1))}^2={(1+2{\alpha }^2-2\alpha )}^2$$$$=4{\alpha }^4-8{\alpha }^3+8{\alpha }^2-4\alpha +1$$$$\because lhs=rhs$$$$\therefore proved$$

Answered by witcher3 last updated on 27/Dec/23

$${\mathrm{x}}_2-{\mathrm{x}}_1=1$$$$\Rightarrow {\mathrm{x}}_2^2+{\mathrm{x}}_1^2-{2\mathrm{x}}_1{\mathrm{x}}_2=1$$$$={\mathrm{x}}_1^2+{\mathrm{x}}_2^2-2\mathrm{m}=1$$$${\mathrm{x}}_1^2+{\mathrm{x}}_2^2={({\mathrm{x}}_1+{\mathrm{x}}_2)}^2-{2\mathrm{x}}_1{\mathrm{x}}_2={\mathrm{l}}^2-2\mathrm{m}$$$$\Rightarrow {\mathrm{l}}^2-2\mathrm{m}-2\mathrm{m}=1$$$$\Rightarrow {\mathrm{l}}^2+{4\mathrm{m}}^2=1+4\mathrm{m}+{4\mathrm{m}}^2={(2\mathrm{m}+1)}^2$$

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