Question

Find the value of a such that the quadratic equation 9x2-3ax+1=0 [ 9 x quare - 3 a x +1 -0] has equal roots​

Answer 1

Quadratic Equation is

$9 {x}^{2} - 3ax + 1 = 0$

Has equal roots

When the quadratic equation has equal roots discriminant is zero, d= 0

$d = {b}^{2} - 4ac\\ {b}^{2} - 4ac = 0 \\ {( - 3a)}^{2} - 4 \times 9 \times 1 = 0 \\ 9 {a}^{2} - 36 = 0 \\ 9 {a}^{2} = 36 \\ a = \sqrt{4} = 2$