Find all values of k for which the equation $(k-2)x² - 2(k-1)x + k = 0$ has equal roots.

Answer: The only value of $k$ for which the equation has equal roots is $k=2$ .

Step-by-step solution

Step 1: Identify coefficients and condition for equal roots

For a quadratic equation in the standard form $ax^2 + bx + c = 0$ , the nature of its roots is determined by the discriminant, $\Delta = b^2 - 4ac$ . If the roots are equal, the discriminant must be zero.

Step 2: Substitute coefficients into the discriminant formula

We identify the coefficients from the given equation: $a = k-2$ , $b = -2(k-1)$ , and $c = k$ . We substitute these values into the discriminant formula $b^2 - 4ac = 0$ .

Step 3: Expand and simplify the equation

First, we expand the squared term and the product of the other two terms. Then, we distribute the 4 and combine like terms. This simplification leads to an unexpected result.

Step 4: Analyze the simplified equation

The equation simplifies to $4 = 0$ , which is a false statement. This means there are no values of $k$ for which the discriminant is zero, given the initial assumption that the equation is quadratic.

Step 5: Consider the case where the equation is not quadratic

A quadratic equation requires the coefficient of $x^2$ to be non-zero. If $k-2 = 0$ , then $k=2$ . In this case, the original equation becomes a linear equation, not a quadratic one.

Step 6: Check the linear case

If $k=2$ , the equation becomes $0x^2 - 2(1)x + 2 = 0$ , which simplifies to $-2x + 2 = 0$ . Solving this linear equation gives $x=1$ . A linear equation has exactly one root, which can be considered 'equal roots' in a degenerate sense.

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Find all values of k for which the equation (k−2)x2−2(k−1)x+k=0(k-2)x² - 2(k-1)x + k = 0(k−2)x2−2(k−1)x+k=0 has equal roots.