JEE MATHEMATICS MCQ: THEORY OF EQUATION

1 Easy

1 of 1

Application +4 | -1

If the roots of the equation

8{x^3} - 14{x^2} + 7x - 1 = 0

are in GP, then the roots are

View answer and explanation

Correct option: A

1,\frac{1}{2},\frac{1}{4}

Option A (

1,\frac{1}{2},\frac{1}{4}

) is correct because the roots are in GP and their product is

\frac{1}{8}

, allowing us to find the common ratio.

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,