Question:

The real number $k$ for which the equation $2x^3 + 3x + k = 0$ has two distinct real roots in $[0, 1]$

Updated On: Feb 14, 2025
  • lies between $2$ and $3$
  • lies between $-1$ and $0$
  • does not exist
  • lies between $1$ and $2$
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The Correct Option is C

Solution and Explanation

Let $f (x) = 2x^3 + 3x + k, f'(x) = 6x^2 + 3 > 0$ Thus $f$ is strictly increasing. Hence it has atmost one real root. But a polynomial equation of odd degree has atleast one root. Thus the equation has exactly one root. Then the two distinct' roots; in any interval whatsoever is an impossibility. No such does not exists.
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Concepts Used:

Quadratic Equations

A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers. 

Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.

The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)

Two important points to keep in mind are:

  • A polynomial equation has at least one root.
  • A polynomial equation of degree ‘n’ has ‘n’ roots.

Read More: Nature of Roots of Quadratic Equation

There are basically four methods of solving quadratic equations. They are:

  1. Factoring
  2. Completing the square
  3. Using Quadratic Formula
  4. Taking the square root