Question

find the set of values of k for which kx^2+x+k-1=0 has two distinct roots.

Answer 1

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.

1. For \( k < \frac{1 - \sqrt{2}}{2} \) (let's say \( k = -1 \)):
\[
4(-1)^2 - 4(-1) - 1 = 4 + 4 - 1 = 7 > 0
\]

2. For \( k \) in \( \left( \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right) \) (let's say \( k = 0 \)):
\[
4(0)^2 - 4(0) - 1 = -1 < 0
\]

3. 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

\[
\boxed{\left( \frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2} \right)}
\]