Question

ex 2..................

Answer 1

$\displaystyle 2a) \frac{2}{x} = \frac{x}{8} \Rightarrow 2 \cdot 8=x \cdot x \Rightarrow 16=x^2 \Rightarrow x^2=16\Rightarrow x= \sqrt{16} \Rightarrow \boxed{x=4}\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow \boxed{x=-4}$

$\displaystyle b) \frac{x-1}{2} = \frac{18}{x-1} \Rightarrow (x-1)(x-1)=2 \cdot 18 \Rightarrow (x-1)^2=36 \Rightarrow \\ \\ \Rightarrow x^2-2x+1=36 \Rightarrow x^2-2x+1-36=0 \Rightarrow x^2-2x-35=0 \\ \\ a=1,~b=-2,~c=-35 \\ \\ \Delta=b^2-4ac=(-2)^2-4\cdot1\cdot (-35)=4+140=144\ \textgreater \ 0 \\ \\ x_1= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{2+\sqrt{144} }{2 \cdot 1} = \frac{2+12}{2} = \frac{14}{2} =\boxed{7}\\ \\x_2=\frac{-b- \sqrt{\Delta} }{2a} = \frac{2- \sqrt{144} }{2 \cdot 1}=\frac{2-12}{2}=\frac{-10}{2} =\boxed{-5}$

$\displaystyle c) \frac{2}{x- \sqrt{3} } = \frac{x- \sqrt{3} }{24} \Rightarrow 2 \cdot 24=\left(x- \sqrt{3} \right)\left(x- \sqrt{3} \right) \Rightarrow \\ \\ \Rightarrow 48=\left(x- \sqrt{3} \right)^2 \Rightarrow 48=x^2-2 \sqrt{3} x+3 \Rightarrow \\ \\ \Rightarrow -x^2+2 \sqrt{3} x-3+48=0 \Rightarrow -x^2+2 \sqrt{3}x +45=0 | \cdot (-1) \Rightarrow \\ \\ \Rightarrow x^2-2 \sqrt{3} x-45=0 \\ \\ a=1,~b=-2 \sqrt{3} ,~c=-45 \\ \\ \Delta=b^2-4ac=\left(-2 \sqrt{3} \right)^2-4 \cdot 1 \cdot (-45)=12+180=192$

$\displaystyle x_1= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{2 \sqrt{3}+ \sqrt{192} }{2 \cdot 1} = \frac{2 \sqrt{3}+8 \sqrt{3} }{2} = \frac{10 \sqrt{3} }{2} =\boxed{5 \sqrt{3} }\\ \\ x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{2 \sqrt{3} - \sqrt{192} }{2 \cdot 1} = \frac{2 \sqrt{3}-8 \sqrt{3} }{2} = \frac{-6 \sqrt{3} }{2} =\boxed{-3 \sqrt{3} }$

$\displaystyle d) \frac{|x|}{3} = \frac{12}{|x|} \Rightarrow |x| \cdot |x|=3 \cdot 12 \Rightarrow |x|^2=36 \Rightarrow x^2=36 \Rightarrow x= \sqrt{36} \Rightarrow \\ \\ \Rightarrow \boxed{x=6} \\ \\ \Rightarrow \boxed{x=-6}$

Answer 2

Eu am rezolvat printr-o metoda de clasa 7a, fara a folosi ecuatia de gradul al doilea.