Find all values of the constant k for which the equation x^2 - (2k - 1)x + 1 = 0 has one root in the interval (0, 1) and the other root in the interval (2, 3).

Step-by-step solution

1. 1

   Translate root location into sign conditions, not into the discriminant. Let $f(x)=x^2-(2k-1)x+1$. The parabola opens upward, so it is negative exactly between its two roots. Requiring one root strictly inside $(0,1)$ and the other strictly inside $(2,3)$ is the same as requiring the four sign conditions

   $f(0)>0,\qquad f(1)<0,\qquad f(2)<0,\qquad f(3)>0.$

   Each strict sign change forces a root between the two consecutive test points, and four such points with pattern $+,-,-,+$ trap exactly one root in $(0,1)$ and one in $(2,3)$. The discriminant condition comes for free — a sign change already guarantees real roots.

2. 2

   Evaluate $f$ at the four test points.

   $f(0)=1$

   $f(1)=1-(2k-1)+1=3-2k$

   $f(2)=4-2(2k-1)+1=7-4k$

   $f(3)=9-3(2k-1)+1=13-6k$

3. 3

   Impose the conditions one at a time.

   $f(0)=1>0\quad\text{— automatically satisfied for every }k.$

   $f(1)=3-2k<0\;\Longrightarrow\;k>\frac32$

   $f(2)=7-4k<0\;\Longrightarrow\;k>\frac74$

   $f(3)=13-6k>0\;\Longrightarrow\;k<\frac{13}{6}$

4. 4

   Intersect the constraints. Since $\tfrac74>\tfrac32$, the condition from $f(1)$ is absorbed by the one from $f(2)$, leaving

   $\frac74<k<\frac{13}{6}.$

5. 5

   Verify with an independent argument using the product of the roots. By Vieta, the product of the roots is $1$, so the two roots are reciprocals: $r$ and $1/r$. If $r\in(2,3)$ then automatically $1/r\in\left(\tfrac13,\tfrac12\right)\subset(0,1)$ — the condition on the small root carries no extra information, exactly as the algebra showed. The sum gives

   $r+\frac1r=2k-1\;\Longrightarrow\;k=\frac12\left(r+\frac1r+1\right),$

   and $r+\tfrac1r$ is increasing for $r>1$, so as $r$ runs over $(2,3)$, $k$ runs over $\left(\tfrac12\left(\tfrac52+1\right),\tfrac12\left(\tfrac{10}{3}+1\right)\right)=\left(\tfrac74,\tfrac{13}{6}\right)$ — the same interval.

6. 6

   Spot-check a value inside and one just outside. For $k=2$: $x^2-3x+1=0$ gives roots $0.382$ and $2.618$ — one in each interval, as required. For $k=2.2$ (just above $13/6\approx2.1667$): roots $0.325$ and $3.075$, and the large root has escaped past $3$, confirming the upper endpoint.