Solve for \( x \) and \( y \). iven \( k x^{2}-5 x-1=0 \); wath \( k \neq 0 \) Fa which value(s) of \( k \) will \( k x^{2}-5 x-1=0 \) have the rod inetermine thoo integral values of \( k \) for which root of \( k x^{2}-5 x-1=0 \) will be rational

To determine the conditions under which the equation \( k x^2 - 5x - 1 = 0 \) has rational roots, we can use the discriminant method. The roots of a quadratic equation \( ax^2 + bx + c = 0 \) are given by the formula: \[ D = b^2 - 4ac \] For rational roots, the discriminant \( D \) must be a perfect square. Here, \( a = k \), \( b = -5 \), and \( c = -1 \). So, the discriminant is: \[ D = (-5)^2 - 4(k)(-1) = 25 + 4k \] Setting \( D \) equal to a perfect square \( m^2 \), we get \( 25 + 4k = m^2 \). Rearranging gives: \[ 4k = m^2 - 25 \quad \Rightarrow \quad k = \frac{m^2 - 25}{4} \] Now, for \( k \) to be integral, \( m^2 - 25 \) must be divisible by 4. Since \( 25 \equiv 1 \mod 4 \), \( m^2 \) must also be congruent to 1 modulo 4, so \( m \) must be odd. Taking odd integer values for \( m \) (like 1, 3, 5, ...) helps us find rational \( k \). For instance, if \( m = 1 \): \[ k = \frac{1^2 - 25}{4} = \frac{-24}{4} = -6 \quad (integral) \] If \( m = 3 \): \[ k = \frac{3^2 - 25}{4} = \frac{-16}{4} = -4 \quad (integral) \] Continuing this process, we can find all odd \( m \) values resulting in rational \( k \), allowing for rational roots in our original quadratic equation. Now, moving on to checking potential values for \( k \)—you could find some additional candidates by continuing this sequence, with odd \( m \) generating various \( k \) values, including \( k = -2 \) or even \( k = 0 \) when considering higher odd values. Always verify if the corresponding \( D \) remains a perfect square!