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Question

The equation whose roots are $-2$ and $3$ is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$x^2 - x - 6 = 0$

Finding the Quadratic Equation from Roots

To find the quadratic equation given its roots, we use the relationship between the roots and the coefficients of the equation. If the roots of a quadratic equation are α and β, the equation can be written as: $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ or $x^2 - (\alpha + \beta)x + \alpha\beta = 0$

Identify Given Roots

The given roots are:

  • Root 1: α = -2
  • Root 2: β = 3

Calculate Sum and Product of Roots

Calculate the sum and product using the given roots:

  • Sum of roots:
    $\alpha + \beta = -2 + 3 = 1$
  • Product of roots:
    $\alpha \beta = (-2) \times 3 = -6$

Form the Quadratic Equation

Substitute the calculated sum and product into the standard equation format:

$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
$x^2 - (1)x + (-6) = 0$
$x^2 - x - 6 = 0$

This equation matches option 2.

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Important Questions from Quadratic Equation

  1. For what values of k, the roots of 9x 2 + 8kx + 16 = 0 are real and equal?

  2. If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of  \(\rm \left( 1+\frac{1}{x} \right) \)  is equal to :
  3. The nature of the roots of the equation 4x 2 - 2x - 3 = 0.

  4. If the roots of the equation (q – r)x 2+ (r – p)x + (p – q) = 0 are equal, then which of the following is true?

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