To determine the solutions to the quadratic equation \(4x^2 + 2x - 10 = 0\), we can rearrange it into standard form and then apply the quadratic formula.
First, we rewrite the equation as: \[ 4x^2 + 2x - 10 = 0 \]
Next, we can apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 4\), \(b = 2\), and \(c = -10\).
Now, we calculate the discriminant: \[ b^2 - 4ac = 2^2 - 4(4)(-10) = 4 + 160 = 164 \]
Now we can substitute back into the quadratic formula: \[ x = \frac{-2 \pm \sqrt{164}}{2 \cdot 4} \] Simplifying the expression gives: \[ x = \frac{-2 \pm \sqrt{164}}{8} \]
Next, we can simplify \(\sqrt{164}\): \[ \sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41} \]
So now we substitute this back into our formula: \[ x = \frac{-2 \pm 2\sqrt{41}}{8} = \frac{-1 \pm \sqrt{41}}{4} \]
Now we have two potential solutions:
- \(x = \frac{-1 + \sqrt{41}}{4}\)
- \(x = \frac{-1 - \sqrt{41}}{4}\)
Using a calculator to approximate these values:
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For \(x = \frac{-1 + \sqrt{41}}{4}\): \[ \sqrt{41} \approx 6.403 \implies x \approx \frac{-1 + 6.403}{4} \approx \frac{5.403}{4} \approx 1.35075 \approx 1.35 \text{ (rounded to the nearest hundredth)} \]
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For \(x = \frac{-1 - \sqrt{41}}{4}\): \[ x \approx \frac{-1 - 6.403}{4} \approx \frac{-7.403}{4} \approx -1.85075 \approx -1.85 \text{ (rounded to the nearest hundredth)} \]
Thus, the solutions to the quadratic equation \(4x^2 + 2x - 10 = 0\) are approximately \(1.35\) and \(-1.85\).