KnightAngwlidn
Answered

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Hey guys pls check my answer if it is correct. The(6x) on the side is the GCF and if i ever got it wrong pls post the right answer. If you think its correct pls comment/say yes" with your supporting details for me to understand further. Thank you ^^

Hey Guys Pls Check My Answer If It Is Correct The6x On The Side Is The GCF And If I Ever Got It Wrong Pls Post The Right Answer If You Think Its Correct Pls Com class=

Sagot :

Riza1idn
[tex]18-x-x^2+13x=0 \\ \\ -x^2+12x+18=0 \\\\ not \\\\ x^2 +12x +18=0 \\\\a=-1 \\===========================\\[/tex]


[tex]3.)\\\\ \frac{ 3}{x} = \frac{ 1+x-13}{6}\\\\x\neq 0 \\\\ \frac{ 3}{x} = \frac{ x-12}{6}\\ \\Means - Extremes \\ Property \ of \ Property \\ If \ \frac{ a}{b} =\frac{c}{d} \ then\ ad=bc [/tex]

[tex]x\cdot (x-12) =3 \cdot 6 \\\\x^2-12x=18 \ \ |\ subtract\ 18\ to\ both\ sides \\\\x^2-12x-18=0 \\ \\ a=1, \ \ b=-12, \ \ c=-18 \\\\x_{1}=\frac{-b-\sqrt{b^2-4\cdot a\cdot c }}{2a} = \frac{12-\sqrt{ (-12)^2-4\cdot 1\cdot (-18) }}{2 } = \frac{12-\sqrt{ 144+72}}{2 } = \\\\=\frac{12-\sqrt{216}}{2 } =\frac{12-\sqrt{36\cdot 6}}{2 } =\frac{12-6\sqrt{ 6}}{2 } =\frac{2(6- 3\sqrt{ 6})}{2 } =6- 3\sqrt{ 10}\\ \\x_{2}=\frac{-b-\sqrt{b^2-4\cdot a\cdot c }}{2a}=\frac{2(6+ 3\sqrt{ 6})}{2 } =6+ 3\sqrt{ 6} [/tex]


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