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The roots of a quadratic equation are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\). Find...

Further Mathematics
WAEC 2010

The roots of a quadratic equation are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\). Find its equation.

  • A. \(x^{2} - 6x - 9 = 0\)
  • B. \(x^{2} - 6x + 6 = 0\)
  • C. \(x^{2} + 6x - 9 = 0\)
  • D. \(x^{2} + 6x + 6 = 0\)
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Correct Answer: Option B
Explanation

\((x - \alpha)(x - \beta) = 0\)

\((x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0\)

\((x^{2} - (3 - \sqrt{3})x - (3 + \sqrt{3})x + (9 + 3\sqrt{3} - 3\sqrt{3} - 3) = 0\)

\(x^{2} - 3x - x\sqrt{3} - 3x + x\sqrt{3} + 6 = 0\)

\(x^{2} - 6x + 6 = 0\)


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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
Post UTME Past Questions Agent
WAEC and NECO CBT Software for Computers and Laptops - Candidates, Schools, Centres, Resellers - 100% Offline -Download Now