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Question

Find the quadratic equation whose roots are equal to the sum and product of the roots of the quadratic equation \(2x^2 + 4x - 1 = 0\).

This question was previously asked in
RRB Group D 2019 Question Paper (26-Aug-2022) (Shift 2)
The correct answer is

\(2x^2 + 5x + 2 = 0\)

For \(2x^2 + 4x - 1 = 0\), sum of roots \(= -\tfrac{b}{a} = -\tfrac{4}{2} = -2\) and product of roots \(= \tfrac{c}{a} = -\tfrac{1}{2}\).

New roots are \(-2\) and \(-\tfrac{1}{2}\); their sum \(= -\tfrac{5}{2}\) and product \(= 1\). The required equation is \(x^2 - (\text{sum})x + \text{product} = 0\), giving \(x^2 + \tfrac{5}{2}x + 1 = 0\).

Multiplying both sides by \(2\): \(2x^2 + 5x + 2 = 0\).

Hence, \(2x^2 + 5x + 2 = 0\) is correct.

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Similar Questions

  1. \(x^2 - 2x + 3 = 0\) is a ______.

  2. One of the roots of the equation \(x^2 - 24x + k = 0\) is \(x = 2\). The other root will be:

  3. If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - 2x - 1 = 0\), then which is the equation whose roots are \(\alpha^2\) and \(\beta^2\)?

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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. 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?

  4. Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is

  5. The value of m for which one of the root of x 2 - 3x + 2m = 0 is double of the root of x 2 - x + m = 0 is:

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