Question:

Let $f(x) = x^2 + ax + b,$ where $a, b \in R$. If $ f(x) = 0$ has all its roots imaginary, then the roots of $ f(x) + f' (x) + f" (x) = 0$ are

Updated On: Jun 23, 2023
  • Real and distinct
  • Imaginary
  • Equal
  • Rational and equal
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The Correct Option is B

Solution and Explanation

Given, $f(x) = x^2 + ax + b $ has imaginary roots.
$ \therefore$ Discriminant, $D < 0 \, \Rightarrow \, a^2 - 4b < 0$
Now, $f'(x) = 2x + a$
$f'(x) = 2$
Also , $f(x) + f (x) + f" (x) = 0 $
$\Rightarrow \, \, \, x^2 + ax + b + 2x + a + 2 = 0 $
$\Rightarrow \, \, \, x^2 + (a + 2)x + b + a + 2 = 0 $
$ \therefore \, \, x = \frac{ - a + 2 \pm \overline{ a+2^2 - 4 \, a + b + 2}}{2}$
$ = \frac{- a + 2 \, \pm \, \overline{a^2 - 4 b - 4}}{2}$
Since, $a^2 - 4b < 0 $
$a^2 - 4b - 4 < 0$
Hence, E (i) has imaginary roots
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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