The graph is that of \(y = 2x^2 - 5x - 3.\) For what value of \(x\) will \(y\) b...
Question
The graph is that of \(y = 2x^2 - 5x - 3.\) For what value of \(x\) will \(y\) be negative? For what value of \(x\) will \(y\) be negative?
Options
A)
\(-\frac{1}{2} \leq x < 3\)
B)
\(-\frac{1}{2} < x \leq 3\)
C)
\(-\frac{1}{2} < x < 3\)
D)
\(-\frac{1}{2} \leq x \leq 3\)
The correct answer is C.
Explanation:
The graph is that of \(y = 2x^2 - 5x - 3.\)
\(2x^2 - 5x - 3 = 0\)
\(2x^2 - 6x + x - 3 = 0\)
\(2x(x - 3) + 1(x - 3) = 0\)
\((2x + 1)(x - 3) = 0\)
\(2x + 1 = 0\)
\(2x = -1\)
\(x = -\frac{1}{2}\)
\(x - 3 = 0\)
\(x = 3\)
For what value of \(x\) will \(y\) be negative?
\(-\frac{1}{2} < x < 3\)
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The graph is that of \(y = 2x^2 - 5x - 3.\)
\(2x^2 - 5x - 3 = 0\)
\(2x^2 - 6x + x - 3 = 0\)
\(2x(x - 3) + 1(x - 3) = 0\)
\((2x + 1)(x - 3) = 0\)
\(2x + 1 = 0\)
\(2x = -1\)
\(x = -\frac{1}{2}\)
\(x - 3 = 0\)
\(x = 3\)
For what value of \(x\) will \(y\) be negative?
\(-\frac{1}{2} < x < 3\)