For what values of y is \(\frac{y + 2}{8y^2 – 10y + 3}\) not defined?
- \(\frac{-3}{4}\), \(\frac{1}{2}\)
- \(\frac{-3}{4}\), \(\frac{-1}{2}\)
-
\(\frac{3}{4}\), \(\frac{1}{2}\)
- \(\frac{3}{4}\), \(\frac{-1}{2}\)
The correct answer is: C
Explanation
\(\frac{y + 2}{8y^2 - 10y + 3}\) not defined when the denominator is equal to zero
8y\(^2\) - 10y + 3 = 0 using factorization method
8y\(^2\) - 6y - 4y + 3 = 0
2y(4y - 3) - 1(4y - 3) = 0
(4y - 3)(2y - 1) = 0
4y - 3 = 0 or 2y - 1 = 0
Hence, y = \(\frac{3}{4}\) or \(\frac{1}{2}\)