Question

For what value of k the quadratic equation kx2 + 11x - 4 = 0 has real roots.

Answer 1

${b}^{2} - 4ac = 0 \\ \\ = > {11}^{2} - 4 \times k \times ( - 4) = 0 \\ \\ = > 121 + 16k = 0 \\ \\ = > 16k = - 121 \\ \\ = > k = \frac{ - 121}{16} \\ \\ = > k = 16 \times \frac{9}{7}$