if α, β, are the roots of equation x³-4 x²+x+6 then the equation roots are \; 1/α \; 1/β, \; 1/ is

Correct Answer is : 6x3+x2−4x+1=06 x³+x²-4 x+1=06x3+x2−4x+1=0

Correct Answer is : 6x3+x2−4x+1=06 x³+x²-4 x+1=06x3+x2−4x+1=0

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Question : 38 of 39

Marks: +1, -0

\alpha, \beta, \gamma

x^{3}-4 x^{2}+x+6

\; \frac{1}{\alpha}

\; \frac{1}{\beta}, \; \frac{1}{\gamma}

Solution:

Concept:

To find an equation whose roots are reciprocals of the roots of a given polynomial, replace

x

\frac{1}{x}

and simplify.

Explanation:

Let

f(x) = x^{3} - 4x^{2} + x + 6 = 0

have roots

\alpha, \beta, \gamma

The required equation has roots

\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}

Substitute

x = \frac{1}{y}

f(x) = 0

\left(\frac{1}{y}\right)^3 - 4\left(\frac{1}{y}\right)^2 + \frac{1}{y} + 6 = 0

Multiply by

y^3

1 - 4y + y^2 + 6y^3 = 0

Rearrange:

6y^3 + y^2 - 4y + 1 = 0

Replace

y

x

6x^3 + x^2 - 4x + 1 = 0

Answer:

The correct equation is

6x^{3} + x^{2} - 4x + 1 = 0

, which corresponds to option C.

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