Answer:
x = 6
x = 0
Step-by-step explanation:
identify the coefficients
[tex]9x^2-54x=0\\a=9\\b=-54\\c=0[/tex]
coefficient a = 1
[tex]9x^2-54x=0\\9/9x^2-54x/9=0/9\\x^2-6x=0\\\rightarrow a=1\\\rightarrow b=-6\\\rightarrow c=0[/tex]
complete the square
[tex]b=-6\\(\frac{b}{2})^2=(\frac{-6}{2})^2\\(\frac{x}{y})^2=\frac{x^2}{y^2}\\=(\frac{-6}{2})^2=\frac{-6^2}{2^2}\\\frac{-6^2}{2^2}=\frac{36}{4}\\36 \div 4=9\\\\\downarrow\\\\x^2-6x=0\\x^2-6x+9=0+9\\x^2-6x+9=9\\\frac{b}{2}\\\frac{b}{2}=\frac{-6}{2}\\\frac{b}{2}=\frac{\left(-3\cdot 2\right)}{\left(1\cdot 2\right)}\\\frac{b}{2}=-3\\x^2-6x+9=9\\(x-3)^2=9\\[/tex]
solve for x
[tex](x-3)^2=9\\\sqrt{(x-3)^2}=\sqrt{9}\\x-3=+-\sqrt{9}\\x-3+3=3+-\sqrt{9}\\x=3+-\sqrt{9}\\x=3+-3\\x_1=6\\x_2=0[/tex]
≡ In the end, you have 2 values for x, because you can either subtract 3 by 3 or add 3 by 3 to get six or zero.
