JEE MATHEMATICS MCQ: THEORY OF EQUATION

1 Easy

1 of 1

Application +4 | -1

\alpha ,\beta

and

\gamma

are the roots of equation

{x^3} - 3{x^2} + x + 5 = 0

, then

y = \sum {\alpha ^2} + \alpha \beta \gamma

satisfies the equation

View answer and explanation

Correct option: B

{y^3} - {y^2} - y - 2 = 0

Option B (

{y^3} - {y^2} - y - 2 = 0

) is correct because the value of y is 2, which satisfies the given equation.

More Questions from THEORY OF EQUATION

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{x^{2n}} - 1 = 0

\frac{1}{x}+\frac{1}{\sqrt{\left(1-x^2\right)}}=\frac{35}{12}

x^7+14 x^5+16 x^3+30 x-560=0

x_1, x_2, x_3, \ldots, x_n

x^n+a x+b=0

\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right)

P(x)=k x^3+3 x^2-3

Q(x)=2 x^3-5 x+k

(x-4)

k

{\left( {x - 1} \right)^3}

{x^4} + a{x^3} + b{x^2} + cx - 1,