Question

1. Find the value of a if-1/2 is zero of f(x) = 6x2 + x + a​

Answer 1

Answer:

a = -1

Step-by-step explanation:

$f(x) = 6 {x}^{2} + x \: + a \\ \\ f( \frac{ - 1}{2} ) = 6( \frac{ - 1}{2})^{2} + ( \frac{ - 1}{2}) + a \\ \\ f( \frac{ - 1}{2} ) = 6 \times \frac{1}{4} - \frac{1}{2} + a \\ \\ f( \frac{ - 1}{2} ) = \frac{3}{2} - \frac{1}{2} + a \\ \\ f( \frac{ - 1}{2} ) = \ \frac{2}{2} + a \\ \\ f( \frac{ - 1}{2} ) = 1 + a \\ \\ \\ \\ \\ </p><p> \\ As \: \frac{ - 1}{2} \: is \: a \: zero \: of \: polynomial \: f(x). \\ Therefore \: \: f( \frac{ - 1}{2}) =0 \\ \\ \\ \\ f( \frac{ - 1}{2}) =0 \\ \\ →1 \: + a \: = 0 \\ →a \: = \: - 1$

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