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 solve for \( x \) and \( y \) in the equation \( k x^{2} - 5 x - 1 = 0 \), we need to determine the conditions under which the roots of this quadratic equation are rational. ### Step 1: Identify the quadratic formula The roots of a quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \( a = k \), \( b = -5 \), and \( c = -1 \). ### Step 2: Calculate the discriminant The discriminant \( D \) is given by: \[ D = b^2 - 4ac \] Substituting the values: \[ D = (-5)^2 - 4(k)(-1) = 25 + 4k \] ### Step 3: Condition for rational roots For the roots to be rational, the discriminant must be a perfect square. Therefore, we need: \[ 25 + 4k = m^2 \] for some integer \( m \). ### Step 4: Rearranging the equation Rearranging gives: \[ 4k = m^2 - 25 \] \[ k = \frac{m^2 - 25}{4} \] ### Step 5: Finding integral values of \( k \) For \( k \) to be an integer, \( m^2 - 25 \) must be divisible by 4. We can analyze \( m^2 - 25 \): \[ m^2 - 25 = (m - 5)(m + 5) \] Both \( m - 5 \) and \( m + 5 \) must be even for their product to be divisible by 4. This occurs when \( m \) is odd. ### Step 6: Finding odd integers for \( m \) Let \( m = 2n + 1 \) for integers \( n \): \[ k = \frac{(2n + 1)^2 - 25}{4} \] Calculating: \[ (2n + 1)^2 = 4n^2 + 4n + 1 \] Thus, \[ k = \frac{4n^2 + 4n + 1 - 25}{4} = n^2 + n - 6 \] ### Step 7: Finding integral values of \( k \) Now we can find integral values of \( k \) by substituting different integer values for \( n \): - For \( n = -3 \): \( k = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0 \) (not valid since \( k \neq 0 \)) - For \( n = -2 \): \( k = (-2)^2 + (-2) - 6 = 4 - 2 - 6 = -4 \) - For \( n = -1 \): \( k = (-1)^2 + (-1) - 6 = 1 - 1 - 6 = -6 \) - For \( n = 0 \): \( k = 0^2 + 0 - 6 = -6 \) - For \( n = 1 \): \( k = 1^2 + 1 - 6 = 1 + 1 - 6 = -4 \) - For \( n = 2 \): \( k = 2^2 + 2 - 6 = 4 + 2 - 6 = 0 \) (not valid since \( k \neq 0 \)) - For \( n = 3 \): \( k = 3^2 + 3 - 6 = 9 + 3 - 6 = 6 \) ### Conclusion The integral values of \( k \) for which the roots of \( k x^{2} - 5 x - 1 = 0 \) will be rational are \( k = -4, -6, 6 \).