To solve the given equation:

$x^2 - 6x - 1 + k(2x + 1) = 0,$

we aim to simplify and solve it for $x$. Here's how:

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Step 1: Expand the equation

Expand $k(2x + 1)$:

$x^2 - 6x - 1 + 2kx + k = 0$

Combine like terms:

$x^2 + (-6 + 2k)x + (-1 + k) = 0.$

The quadratic equation becomes:

$x^2 + (-6 + 2k)x + (-1 + k) = 0.$

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Step 2: Solve using the quadratic formula

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$

where:

• $a = 1$,
• $b = -6 + 2k$,
• $c = -1 + k$.

Substitute these into the formula:

$x = \frac{-(-6 + 2k) \pm \sqrt{(-6 + 2k)^2 - 4(1)(-1 + k)}}{2(1)}.$

Simplify further:

$x = \frac{6 - 2k \pm \sqrt{(-6 + 2k)^2 - 4(-1 + k)}}{2}.$

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Step 3: Simplify the discriminant

The discriminant ($D$) is:

$D = (-6 + 2k)^2 - 4(1)(-1 + k).$

Expand $(-6 + 2k)^2$:

$(-6 + 2k)^2 = 36 - 24k + 4k^2.$

Simplify $4(-1 + k)$:

$4(-1 + k) = -4 + 4k.$

Combine:

$D = 36 - 24k + 4k^2 + 4 - 4k.$

$D = 40 - 28k + 4k^2.$

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Final Quadratic Solution:

The solution becomes:

$x = \frac{6 - 2k \pm \sqrt{40 - 28k + 4k^2}}{2}.$

This represents the roots of the equation in terms of $k$.

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Would you like me to:

1. Explore the conditions for real roots (discriminant $D \geq 0$)?
2. Solve for specific $k$ values?
3. Analyze symmetry or special cases?
4. Graph the equation?
5. Verify solutions for specific numerical inputs?

Tip: Always check the discriminant ($D$) when solving quadratics—it determines the nature of the roots!