If the roots of the equation \(x^2-(k-2)x + (k + 1) = 0\) are equal, then what are the values of \(k\)?
Understanding Equal Roots in Quadratic Equations
The question asks us to find the values of the parameter 'k' for the quadratic equation \(x^2 - (k-2)x + (k+1) = 0\) such that its roots are equal.
Quadratic Equation Basics
A standard quadratic equation is given in the form \(ax^2 + bx + c = 0\).
In our given equation, \(x^2 - (k-2)x + (k+1) = 0\), we can identify the coefficients:
- \(a = 1\)
- \(b = -(k-2)\)
- \(c = (k+1)\)
The Condition for Equal Roots
For a quadratic equation \(ax^2 + bx + c = 0\), the nature of its roots is determined by the discriminant, denoted by \(\Delta\) or \(D\). The formula for the discriminant is:
\( \Delta = b^2 - 4ac \)
The conditions based on the discriminant are:
- If \(\Delta > 0\), the equation has two distinct real roots.
- If \(\Delta = 0\), the equation has exactly one real root, or two equal real roots.
- If \(\Delta < 0\), the equation has two complex conjugate roots.
Since the question states that the roots are equal, we must have the discriminant equal to zero:
\( \Delta = 0 \)
Calculating the Discriminant
Now, let's substitute the coefficients \(a\), \(b\), and \(c\) from our specific equation into the discriminant formula:
\( \Delta = (-(k-2))^2 - 4(1)(k+1) \)
Solving for k
We set the discriminant to zero and solve for 'k':
\( (-(k-2))^2 - 4(1)(k+1) = 0 \)
Simplify the equation:
\( (k-2)^2 - 4(k+1) = 0 \)
Expand the terms:
\( (k^2 - 2(k)(2) + 2^2) - (4k + 4) = 0 \)
\( (k^2 - 4k + 4) - 4k - 4 = 0 \)
Combine like terms:
\( k^2 - 4k - 4k + 4 - 4 = 0 \)
\( k^2 - 8k = 0 \)
Factor out the common term 'k':
\( k(k - 8) = 0 \)
This equation gives two possible values for 'k':
- \(k = 0\)
- \(k - 8 = 0 \implies k = 8\)
Therefore, the values of 'k' for which the roots of the equation \(x^2 - (k-2)x + (k+1) = 0\) are equal are 0 and 8.
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