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Find the quadratic equation whose roots are \(\frac{2}{3} and \frac{- 3}{4}\)

Mathematics
WAEC 2023

Find the quadratic equation whose roots are \(\frac{2}{3} and \frac{- 3}{4}\)

  • A. \(12y^2 - y - 6 = 0\)
  • B. \(12y^2 - y + 6 = 0\)
  • C. \(12y^2 + y - 6 = 0\)
  • D. \(y^2 + y - 6 = 0\)
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Correct Answer: Option C
Explanation

Let p = \(\frac{2}{3}\) and q =  \(\frac{- 3}{4}\)

using (y - p)(y - q) = 0

= ( y -  \(\frac{2}{3})\)( y - (\(\frac{- 3}{4})) = 0\)

 =  (\( y -  \frac{2}{3})( y + \frac{3}{4})\) = 0

\( y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{6}{12} = 0 \)

\( y^2 + \frac{1}{12}y - \frac{1}{2}\) = 0

= multiply through by the l. c. m of 3 and 4 = 12

∴ the quadratic equation is  \(12y^2 + y - 6 = 0\)

 


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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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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free