To solve the equation \(2x^2 + x - 5 = 0\) and find the zeros, we will use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, we have:
- \(a = 2\)
- \(b = 1\)
- \(c = -5\)
Now, we will substitute these values into the quadratic formula:
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Calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 - 4ac = 1^2 - 4(2)(-5) = 1 + 40 = 41 \]
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Find the two solutions using the quadratic formula: \[ x = \frac{-1 \pm \sqrt{41}}{2 \cdot 2} = \frac{-1 \pm \sqrt{41}}{4} \]
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Calculate the two possible values for \(x\):
- For the positive root: \[ x_1 = \frac{-1 + \sqrt{41}}{4} \]
- For the negative root: \[ x_2 = \frac{-1 - \sqrt{41}}{4} \]
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Next, we can approximate these values. First, we find \(\sqrt{41}\): \[ \sqrt{41} \approx 6.403 \]
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Calculate \(x_1\) and \(x_2\):
- For \(x_1\): \[ x_1 = \frac{-1 + 6.403}{4} \approx \frac{5.403}{4} \approx 1.35075 \]
- For \(x_2\): \[ x_2 = \frac{-1 - 6.403}{4} \approx \frac{-7.403}{4} \approx -1.85075 \]
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Rounding to the hundredths place, we get:
- \(x_1 \approx 1.35\)
- \(x_2 \approx -1.85\)
Thus, the solutions, listed in increasing order, are: \[ x = -1.85, \quad 1.35 \]