Roots of the equation

Algebra Level 3

If the two roots of the equation 3 x 2 − 18 x + 2 = 0 3x^2 - 18x + 2 =0 are − α 2 β -\dfrac{\alpha^2}{\beta} and − β 2 α -\dfrac{\beta^2}{\alpha} , then determine which of these equations has roots by α \alpha and β \beta .

x 2 − 6 x + 2 = 0 x^2-6x+2=0 3 x 2 − 6 x + 2 = 0 3x^2-6x+2=0 2 x 2 − 3 x + 2 = 0 2x^2-3x+2=0 3 x 2 + 6 x + 2 = 0 3x^2+6x+2=0

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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1 solution

Anirudh Sreekumar
Oct 17, 2017

α β = 2 3 α 2 β + β 2 α = − 6 α 3 + β 3 = − 6 α β = − 4 let p = α + β p 3 = 3 α β p + α 3 + β 3 p 3 = 2 p − 4 ⟹ p 3 − 2 p + 4 = 0 ( p + 2 ) ( p 2 − 2 p + 2 ) = 0 since p is real and p 2 − 2 p + 2 doesnot have any real roots, The only possible value for p is, p = − 2 ⟹ The equation should be, 3 x 2 + 6 x + 2 \begin{aligned}\alpha\beta&=\dfrac{2}{3}\\ \dfrac{\alpha^2}{\beta}+\dfrac{\beta^2}{\alpha}&=-6\\ \alpha^3+\beta^3&=-6\alpha\beta=-4\\\\ \text{let } p=\alpha&+\beta\\\\ p^3&=3\alpha\beta p+\alpha^3+\beta^3\\ p^3&=2p-4\\ \implies p^3-2p+4&=0\\ (p+2)(p^2-2p+2)&=0\\ \text{since p is real and }&p^2-2p+2 \text{ doesnot have any real roots,}\\ \text{The only possible } &\text{value for p is, }p=-2\\ \implies \text{The equation should be,}& 3x^2+6x+2\end{aligned}

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