To determine the values of \( k \) for which the quadratic equation
\[ kx^2 + x + (k-1) = 0 \]
has two distinct roots, we must analyze the discriminant of the equation. The general form of a quadratic equation is given by:
\[ ax^2 + bx + c = 0 \]
where in this case, \( a = k \), \( b = 1 \), and \( c = k-1 \).
The discriminant \( D \) of a quadratic equation is given by:
\[ D = b^2 - 4ac \]
For the quadratic to have two distinct roots, the discriminant must be positive:
\[ D > 0 \]
Now, substituting for \( a \), \( b \), and \( c \) in our equation, we compute the discriminant:
\[ D = 1^2 - 4(k)(k - 1) = 1 - 4k(k - 1) \]
Simplifying this gives:
\[ D = 1 - 4k^2 + 4k \]
We want this to be greater than zero:
\[ 1 - 4k^2 + 4k > 0 \]
Rearranging this inequality, we have:
\[ -4k^2 + 4k + 1 > 0 \]
Multiplying through by \(-1\) (and reversing the inequality) gives:
\[ 4k^2 - 4k - 1 < 0 \]
Next, we solve the quadratic equation \( 4k^2 - 4k - 1 = 0 \) to find the critical points. Using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 4 \), \( b = -4 \), and \( c = -1 \):
\[ k = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \] \[ k = \frac{4 \pm \sqrt{16 + 16}}{8} \] \[ k = \frac{4 \pm \sqrt{32}}{8} \] \[ k = \frac{4 \pm 4\sqrt{2}}{8} \] \[ k = \frac{1 \pm \sqrt{2}}{2} \]
The two critical points are:
\[ k_1 = \frac{1 - \sqrt{2}}{2}, \quad k_2 = \frac{1 + \sqrt{2}}{2} \]
To determine the intervals for which \( 4k^2 - 4k - 1 < 0 \), we test the signs of the quadratic expression in the intervals defined by these critical points.
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For \( k < \frac{1 - \sqrt{2}}{2} \) (let's say \( k = -1 \)): \[ 4(-1)^2 - 4(-1) - 1 = 4 + 4 - 1 = 7 > 0 \]
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For \( k \) in \( \left( \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right) \) (let's say \( k = 0 \)): \[ 4(0)^2 - 4(0) - 1 = -1 < 0 \]
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For \( k > \frac{1 + \sqrt{2}}{2} \) (let's say \( k = 2 \)): \[ 4(2)^2 - 4(2) - 1 = 16 - 8 - 1 = 7 > 0 \]
From this analysis, we find that the quadratic \( 4k^2 - 4k - 1 < 0 \) is satisfied in the interval:
\[ \left( \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right) \]
Thus, the set of values of \( k \) for which the quadratic equation \( kx^2 + x + (k-1) = 0 \) has two distinct roots is
\[ \boxed{\left( \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right)} \]
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