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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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Similar Questions

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  7. Find the least positive integer such that its square is greater than 5 times of the integer by $-6$.
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Important Questions from Quadratic Equation

  1. If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\)  is:

  2. If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:

  3. If x 2 – 3x + 1 = 0, then the value of  \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\)  is:

  4. If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) \(x \ne 0\) , then what is the value of  \((x^4+{1\over{x^2}})\over(x^2+1) \)  ?

  5. If x 2\(\frac{1}{x^2}\)  = 18, x > 0, then find the value of x \(\frac{1}{x^3}\) .

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