To solve $kx^2 + 2x + k = 0$ by completing the square, follow these steps:

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Step 1: Factor out $k$ from the quadratic and constant terms

$k\left(x^2 + \frac{2}{k}x + \frac{k}{k}\right) = 0$ $k\left(x^2 + \frac{2}{k}x + 1\right) = 0$

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Step 2: Prepare to complete the square

The coefficient of $x$ is $\frac{2}{k}$. Half of this is $\frac{1}{k}$, and squaring it gives $\left(\frac{1}{k}\right)^2 = \frac{1}{k^2}$.

To complete the square, add and subtract $\frac{1}{k^2}$ inside the parentheses: $k\left(x^2 + \frac{2}{k}x + \frac{1}{k^2} - \frac{1}{k^2} + 1\right) = 0$

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Step 3: Rewrite as a perfect square

The first three terms form a perfect square: $k\left(\left(x + \frac{1}{k}\right)^2 - \frac{1}{k^2} + 1\right) = 0$ Simplify the constant terms: $k\left(\left(x + \frac{1}{k}\right)^2 + \left(1 - \frac{1}{k^2}\right)\right) = 0$

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Step 4: Solve for $x$

Expand $1 - \frac{1}{k^2}$: $k\left(\left(x + \frac{1}{k}\right)^2 + \frac{k^2 - 1}{k^2}\right) = 0$ Divide through by $k$: $\left(x + \frac{1}{k}\right)^2 + \frac{k^2 - 1}{k^2} = 0$ Isolate the perfect square: $\left(x + \frac{1}{k}\right)^2 = -\frac{k^2 - 1}{k^2}$ Take the square root of both sides: $x + \frac{1}{k} = \pm\sqrt{-\frac{k^2 - 1}{k^2}}$ $x + \frac{1}{k} = \pm i\sqrt{\frac{k^2 - 1}{k^2}}$ Simplify: $x = -\frac{1}{k} \pm i\frac{\sqrt{k^2 - 1}}{k}$

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Final Solution:

$x = -\frac{1}{k} \pm i\frac{\sqrt{k^2 - 1}}{k}$

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Do you want details or have any questions? Here are 5 related problems you can explore:

1. Solve $2x^2 + 4x + 3 = 0$ by completing the square.
2. Solve $3x^2 + 6x + 7 = 0$ and express the solution in terms of $i$.
3. Verify the solution of $x^2 + 6x + 10 = 0$ using the quadratic formula and completing the square.
4. Derive the general formula for solving $ax^2 + bx + c = 0$ by completing the square.
5. Explore how completing the square relates to the vertex form of a quadratic equation.

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Tip:

When completing the square, always ensure the coefficient of $x^2$ is 1 before adding and subtracting the necessary term.