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find the solutions(s) to 9x^2-54x+0

Sagot :

beinggreat78

Answer:

x = 6

x = 0

  • As shown, there are 2 values for x. The question is why, which we will go over in the explanation part.

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.

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