If α+β are the roots of the equation 3x^2-5x+6=0. Find the value of α/β+β/α
Let α\alphaα and β\betaβ be roots of the equation 3x2−5x+6=03x^2-5x+6=03x2−5x+6=0 ,
Using Viet Theorem, we obtain that α+β=53\alpha +\beta =\frac{5}{3}α+β=35 and αβ=63=2\alpha \beta=\frac{6}{3}=2αβ=36=2 .
αβ+βα=α2+β2αβ=α2+2αβ+β2−2αβαβ=(α+β)2−2αβαβ=(53)2−2⋅22=25−3618=−1118\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}= \frac{\alpha^2+2\alpha\beta+\beta^2-2\alpha \beta}{\alpha\beta}= \frac{(\alpha+\beta)^2-2\alpha \beta}{\alpha\beta}=\frac{\left(\frac{5}{3}\right)^2-2\cdot 2}{2}=\frac{25-36}{18}=-\frac{11}{18}βα+αβ=αβα2+β2=αβα2+2αβ+β2−2αβ=αβ(α+β)2−2αβ=2(35)2−2⋅2=1825−36=−1811
Answer: αβ+βα=−1118\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=-\frac{11}{18}βα+αβ=−1811 .
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The term 2βα was added, later the term 2βα was subtracted because 0=2βα- 2βα.
Where did you get the -2βα from
Comments
The term 2βα was added, later the term 2βα was subtracted because 0=2βα- 2βα.
Where did you get the -2βα from