Question

If $\alpha,\beta,\gamma$ are real roots of the equation x³-3px²+3qx-1=0, the centroid of the triangle whose vertices are [tex]\bigg(\alpha,\frac{1}{\alpha}\bigg), \bigg(\beta,\frac{1}{\beta}\bigg), \bigg(\gamma,\frac{1}{\gamma}\bigg) is?

a)(p,p)

b) (0,p)

c)(p,q)

d)(q,0)​

Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

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Given:

Equation

x³-3px²+3qx-1 =0 has three Roots $\alpha,\beta\gamma$

Vertices of triangle are:

$\bigg(\alpha,\frac{1}{\alpha}\bigg), \bigg(\beta,\frac{1}{\beta}\bigg), \bigg(\gamma,\frac{1}{\gamma}\bigg)$

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Formula

Centroid of a Triangle = $(x,y) = \dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}$

$\bold{Given} \begin{cases}\tt{\alpha+\beta+\gamma=\dfrac{-b}{a}} \\ \tt{\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}}\\ \tt{\alpha\beta\gamma=\dfrac{-d}{a}}\end{cases}$

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SoluTion:

From The Equation and Formula given

$\alpha+\beta+\gamma=3p \\ \alpha\beta+\beta\gamma+\gamma\alpha=3q \\ \alpha\beta\gamma=1$

Now We can Calculate the Centroid of triangle

We have been Given The vertices as

$x_1=\alpha,x_2=\beta,x_3=\gamma \\ y_1=\dfrac{1}{\alpha},y_2=\dfrac{1}{\beta},y_3=\dfrac{1}{\gamma}$

By putting these Values In the Formula

$(x,y) \implies \dfrac{\alpha+\beta+\gamma}{3},\dfrac{\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}}{3} \\ \\ \implies \dfrac{\cancel{3}p}{\cancel{3}} , \dfrac{\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}}{3} \\ \\ \implies p,\dfrac{\dfrac{\cancel{3}q}{1}}{\cancel{3}} \\ \\ \leadsto{ \boxed{\boxed{p,q}}}$

Answer 2

Answer:

If $\alpha,\beta,\gamma$ are real roots of the equation x³-3px²+3qx-1=0, the centroid of the triangle whose vertices are [tex]\bigg(\alpha,\frac{1}{\alpha}\bigg), \bigg(\beta,\frac{1}{\beta}\bigg), \bigg(\gamma,\frac{1}{\gamma}\bigg) is?

a)(p,p)

b) (0,p)

c)(p,q)✔️✔️✔️✔️

d)(q,0)