Question

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

Answer 1

Given: $\frac{1}{9}{x}^2-\frac{2}{3}x+1=0$
On multiplying both sides by 9, we get:

$$\begin{gathered} 9\left[\frac{1}{9}{x}^2-\frac{2}{3}x+1\right]=0 \\ =>x^2-6x+9=0 \end{gathered}$$

On splitting the middle term –6x as – 3x – 3x, we get:
x² – 3x – 3x + 9 = 0
=> x(x – 3) – 3(x – 3) = 0
=> (x – 3) (x – 3) = 0
We know that if the product of two numbers is zero then at least one of them must be zero.
Thus, x – 3 = 0 or x – 3 = 0
=> x = 3 or x = 3
Therefore, the solution of the given equation is x = 3, 3.