Question:

The roots $\alpha$ and $\beta$ of the quadratic equation $px^2 + qx + r = 0$ are real and opposite signs. The roots of $\alpha(x - \beta^2)+ \beta (x - \alpha)^2 = 0$ are

Updated On: Jun 23, 2023
  • positive
  • negative
  • real and of opposite signs
  • Imaginary
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The Correct Option is C

Solution and Explanation

$\alpha + \beta = - \frac{q}{p} ; \alpha \beta = \frac{r}{p} $ .Clearly $\alpha \beta < 0.$ $\alpha (x - \beta )^2 + \beta (x -\alpha)^2 = 0$ $ \Rightarrow \, (\alpha + \beta)x^2 - 4 \, \alpha \beta x + \alpha \beta (\alpha + \beta) = 0$ Product of its roots = $\alpha \beta < 0$ $\therefore $ roots are real and of opposite signs.
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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