To solve the quadratic equation $4x^2 + 10x - 3 = 0$ by completing the square, follow these steps:

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Step 1: Divide through by the coefficient of $x^2$

The coefficient of $x^2$ is 4, so divide the entire equation by 4 to simplify: $x^2 + \frac{10}{4}x - \frac{3}{4} = 0$ Simplify further: $x^2 + \frac{5}{2}x - \frac{3}{4} = 0$

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Step 2: Rearrange the equation

Move the constant term to the other side: $x^2 + \frac{5}{2}x = \frac{3}{4}$

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Step 3: Complete the square

To complete the square, take half the coefficient of $x$, square it, and add it to both sides. The coefficient of $x$ is $\frac{5}{2}$, so: $\left( \frac{\frac{5}{2}}{2} \right)^2 = \left( \frac{5}{4} \right)^2 = \frac{25}{16}$ Add $\frac{25}{16}$ to both sides: $x^2 + \frac{5}{2}x + \frac{25}{16} = \frac{3}{4} + \frac{25}{16}$

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Step 4: Simplify both sides

The left-hand side becomes a perfect square trinomial: $\left( x + \frac{5}{4} \right)^2$ On the right-hand side, find a common denominator to combine the fractions: $\frac{3}{4} = \frac{12}{16}, \quad \frac{3}{4} + \frac{25}{16} = \frac{12}{16} + \frac{25}{16} = \frac{37}{16}$ Thus: $\left( x + \frac{5}{4} \right)^2 = \frac{37}{16}$

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Step 5: Solve for $x$

Take the square root of both sides: $x + \frac{5}{4} = \pm \sqrt{\frac{37}{16}}$ Simplify: $x + \frac{5}{4} = \pm \frac{\sqrt{37}}{4}$

Subtract $\frac{5}{4}$ from both sides: $x = -\frac{5}{4} \pm \frac{\sqrt{37}}{4}$

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Step 6: Combine terms

Combine into a single fraction: $x = \frac{-5 \pm \sqrt{37}}{4}$

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Final Answer:

$x = \frac{-5 + \sqrt{37}}{4} \quad \text{or} \quad x = \frac{-5 - \sqrt{37}}{4}$

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Would you like a breakdown of any specific step?
Here are some related questions you might want to explore:

1. How is completing the square related to the quadratic formula?
2. What are the applications of completing the square in real-world problems?
3. Why is $\frac{-5 \pm \sqrt{37}}{4}$ in simplest form?
4. Can this method be used if the quadratic equation is already in vertex form?
5. How would the graph of this quadratic equation look?

Tip: Always ensure the coefficient of $x^2$ is 1 before completing the square—it simplifies the process significantly!