Question

Please help me in finding the roots of.

X^2/9-2x/3+1=0

Answer 1

$\frac{ {x}^{2} }{9} - \frac{2x}{3} + 1 = 0 \\ \\ \implies \: \frac{ {x}^{2} - 6x + 9 }{9} = 0 \\ \\ \implies \: {x}^{2} - 6x + 9 = 0$

x²-3x-3x+9 = 0

x(x-3)-3(x-3)

=> x = 3, x = 3

The required roots are 3 and 3.

Both the roots are real and like .

So , the discriminant is = 0.

Discriminant = b²-4ac

Answer 2

$\frac{ {x}^{2} }{9} - \frac{2x}{3} + 1 = 0 \\ {x}^{2} - 6x + 9 = 0 \\ {(x)}^{2} - 2.x.3 + {(3)}^{2} = 0 \\ {(x - 3)}^{2} = 0 \\ x - 3 = 0 \\ x = 3$