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.
\(y^2 + y - 6 = 0\)
Correct answer is C
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 \(12y^2 + y - 6 = 0\)
