Answer: (a) \(x\lt \frac15\) or \(x\gt 6\). (b) The discriminant is always positive.

Answer: (a) \(x\lt \frac15\) or \(x\gt 6\). (b) The discriminant is always positive.

(a) The critical values are found from

\(5x-1=0\quad\text{and}\quad 6-x=0\),

so \(x=\frac15\) and \(x=6\).

The product \((5x-1)(6-x)\) is negative outside these two critical values, so

\(x\lt \frac15\quad\text{or}\quad x\gt 6\).

(b) For the quadratic equation

\((2k+1)x^2-4kx+2k-1=0\),

the discriminant is

\(\Delta=(-4k)^2-4(2k+1)(2k-1)\).

Since

\((2k+1)(2k-1)=4k^2-1\),

we get

\(\Delta=16k^2-4(4k^2-1)=4\).

Thus \(\Delta\gt 0\), so the equation has distinct real roots. Also, \(k\ne-\frac12\) ensures that the coefficient of \(x^2\) is not zero, so the equation is genuinely quadratic.