To solve the quadratic inequality \( 2x^2 - x - 3 \leq 0 \), we first find the roots of the corresponding quadratic equation \( 2x^2 - x - 3 = 0 \).
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = -1 \), and \( c = -3 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \]
Now we apply the quadratic formula:
\[ x = \frac{-(-1) \pm \sqrt{25}}{2 \cdot 2} = \frac{1 \pm 5}{4} \]
This gives us two solutions:
- \( x = \frac{1 + 5}{4} = \frac{6}{4} = 1.5 \)
- \( x = \frac{1 - 5}{4} = \frac{-4}{4} = -1 \)
So the roots of the equation \( 2x^2 - x - 3 = 0 \) are \( x = -1 \) and \( x = 1.5 \).
Next, we can analyze the sign of the quadratic expression \( 2x^2 - x - 3 \) on the intervals determined by these roots:
- \( (-\infty, -1) \)
- \( (-1, 1.5) \)
- \( (1.5, \infty) \)
To determine where the inequality is satisfied, we test points from each interval:
- For \( x = -2 \) in \( (-\infty, -1) \):
\[ 2(-2)^2 - (-2) - 3 = 8 + 2 - 3 = 7 > 0 \quad \text{(not in the solution set)} \]
- For \( x = 0 \) in \( (-1, 1.5) \):
\[ 2(0)^2 - (0) - 3 = -3 < 0 \quad \text{(in the solution set)} \]
- For \( x = 2 \) in \( (1.5, \infty) \):
\[ 2(2)^2 - (2) - 3 = 8 - 2 - 3 = 3 > 0 \quad \text{(not in the solution set)} \]
Now we include the points where the inequality is equal to zero, which are the roots \( x = -1 \) and \( x = 1.5 \).
Thus, the solution set to the inequality \( 2x^2 - x - 3 \leq 0 \) is:
\[ [-1, 1.5] \]
In inequality notation, the solution set is: \[ [-1, 1.5] \]
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