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Question Number 215837 by MATHEMATICSAM last updated on 19/Jan/25

$$\mathrm{If}{b}^3+a^{2}c+{ac}^2=3abc\mathrm{then}\mathrm{prove}\mathrm{that}$$$$\mathrm{one}\mathrm{root}\mathrm{of}{ax}^2+bx+c=0\mathrm{is}\mathrm{the}\mathrm{square}$$$$\mathrm{of}\mathrm{the}\mathrm{other}\mathrm{one}.$$

Answered by MrGaster last updated on 19/Jan/25

$$\mathrm{given}{b}^3+a^{2}c+{ac}^2=3abc$$$$\beta be\mathrm{the}\mathrm{roots}\mathrm{of}{ax}^2+bx+c=0$$$$\mathrm{by}{\mathrm{Vieta}}^{\prime }s\mathrm{formulas}:\mathrm{\Sigma }\alpha =-\frac{b}{a}\wedge \mathrm{\Pi }\alpha =\frac{c}{a}$$$$\mathrm{if}\mathrm{one}\mathrm{root}\mathrm{is}\mathrm{the}◻\mathrm{of}\mathrm{the}\mathrm{other},\mathrm{then}\mathrm{\exists }\gamma :\alpha ={\gamma }^2\wedge \beta =\gamma \vee \alpha =\gamma \wedge \beta ={\gamma }^2$$$$\mathrm{without}\mathrm{loss}\mathrm{of}\mathrm{generality},\mathrm{assume}\alpha ={\gamma }^2\wedge \beta =\gamma$$$$\therefore \mathrm{\Sigma }\alpha ={\gamma }^2+\gamma =-\frac{b}{a}$$$$\mathrm{\Pi }\alpha ={\gamma }^3=\frac{c}{a}$$$$\mathrm{from}\mathrm{beth}\mathrm{sides}\mathrm{by}{a}^3:{\left(\frac{b}{a}\right)}^3+\left(\frac{\mathrm{c}}{a}\right)+{\left(\frac{\mathrm{c}}{a}\right)}^2=3\left(\frac{b}{a}\right)\left(\frac{c}{a}\right)$$$$\mathrm{substitute}\mathrm{\Sigma }\alpha \mathrm{and}\mathrm{\Pi }\alpha :{(-\mathrm{\Sigma }\alpha )}^3+\mathrm{\Pi }\alpha +{\left(\mathrm{\Pi }\alpha \right)}^2=3(-\mathrm{\Sigma }\alpha )\left(\mathrm{\Pi }\alpha \right)$$$${(-({\gamma }^2+\gamma ))}^3+{\gamma }^3+{\left({\gamma }^3\right)}^2=3(-({\gamma }^2+\gamma ))\left({\gamma }^3\right)$$$$-3{\gamma }^4=-3{\gamma }^4$$$$bx+c=0\mathrm{is}\mathrm{indeed}\mathrm{the}◻\mathrm{of}\mathrm{the}\mathrm{other}.$$$$\mathrm{ps}:‘‘{◻}^{″}\mathrm{denotes}\mathrm{squaring}.$$

Commented by A5T last updated on 19/Jan/25

$$\mathrm{From}\mathrm{the}\mathrm{wording}\mathrm{of}\mathrm{the}\mathrm{question},\mathrm{what}\mathrm{is}$$$$\mathrm{required}\mathrm{to}\mathrm{prove}\mathrm{is}:$$$${\mathrm{b}}^3+{\mathrm{a}}^2\mathrm{c}+{\mathrm{ac}}^2=3\mathrm{abc}\Rightarrow {\mathrm{x}}_1=\gamma \wedge {\mathrm{x}}_2={\gamma }^2$$$$\mathrm{You}\mathrm{showed}\mathrm{the}\mathrm{converse}\mathrm{here}:$$$${\mathrm{x}}_1=\gamma \wedge {\mathrm{x}}_2={\gamma }^2\Rightarrow {\mathrm{b}}^3+{\mathrm{a}}^2\mathrm{c}+{\mathrm{ac}}^2=3\mathrm{abc}$$$$\mathrm{Both}\mathrm{are}\mathrm{not}\mathrm{quite}\mathrm{the}\mathrm{same}.$$$$\mathrm{A}\Rightarrow \mathrm{B}\not\equiv \mathrm{B}\Rightarrow \mathrm{A}.$$

Answered by A5T last updated on 19/Jan/25

$$\mathrm{Let}\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0\mathrm{be}{\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2$$$${\mathrm{x}}_1+{\mathrm{x}}_2=\frac{-\mathrm{b}}{\mathrm{a}};{\mathrm{x}}_1{\mathrm{x}}_2=\frac{\mathrm{c}}{\mathrm{a}}$$$${\mathrm{b}}^3+{\mathrm{a}}^2\mathrm{c}+{\mathrm{ac}}^2=3\mathrm{abc}\stackrel{/{\mathrm{a}}^3}{\Rightarrow }{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^3+\left(\frac{\mathrm{c}}{\mathrm{a}}\right)+{\left(\frac{\mathrm{c}}{\mathrm{a}}\right)}^2=3\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{c}}{\mathrm{a}}\right)$$$$\Rightarrow -{({\mathrm{x}}_1+{\mathrm{x}}_2)}^3+{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_1^2{\mathrm{x}}_2^2=-3({\mathrm{x}}_1+{\mathrm{x}}_2)\left({\mathrm{x}}_1{\mathrm{x}}_2\right)$$$$\Rightarrow -({\mathrm{x}}_1^3+{3\mathrm{x}}_1^2{\mathrm{x}}_2+{3\mathrm{x}}_1{\mathrm{x}}_2^2+{\mathrm{x}}_2^3)+{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_1^2{\mathrm{x}}_2^2=-{3\mathrm{x}}_1^2{\mathrm{x}}_2-{3\mathrm{x}}_1{\mathrm{x}}_2^2$$$$\Rightarrow -{\mathrm{x}}_1^3-{\mathrm{x}}_2^3+{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_1^2{\mathrm{x}}_2^2=-{\mathrm{x}}_1^2({\mathrm{x}}_1-{\mathrm{x}}_2^2)+{\mathrm{x}}_2({\mathrm{x}}_1-{\mathrm{x}}_2^2)=0$$$$\Rightarrow ({\mathrm{x}}_1^2-{\mathrm{x}}_2)({\mathrm{x}}_1-{\mathrm{x}}_2^2)=0$$$$\Rightarrow {\mathrm{x}}_1^2={\mathrm{x}}_2\mathrm{or}{\mathrm{x}}_2^2={\mathrm{x}}_1◼$$

Commented by ArshadS last updated on 19/Jan/25

$$5thline:$$$$\Rightarrow -({\mathrm{x}}_1^3+{3\mathrm{x}}_1^2{\mathrm{x}}_2+{3\mathrm{x}}_1{\mathrm{x}}_2^2+{\mathrm{x}}_2^3)+{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_1^2{\mathrm{x}}_2^2=-{3\mathrm{x}}_1^2{\mathrm{x}}_2-{3\mathrm{x}}_1{\mathrm{x}}_2^2$$

Commented by A5T last updated on 19/Jan/25

$$\mathrm{It}\mathrm{was}\mathrm{a}\mathrm{typo},\mathrm{thanks}.$$

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