JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

The number of real roots of the equation

{\left( {x + \frac{1}{x}} \right)^3} + x + \frac{1}{x} = 0

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Correct option: A

Option A (0) is correct because the equation has no real roots since the only possible solution leads to

{x^2} = - 1

, which has no real solutions.

More Questions from QUADRATIC EQUATIONS

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\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

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