Question

$(x*1+ \sqrt{3} )^2- 10*1= x^{2} *1-(2* \frac{ \sqrt{3} }{2} + \sqrt{8} * \frac{ \sqrt{2} }{2} )^2$

Rezolvati ! urgent

Answer 1

$\displaystyle \it (x \cdot 1+ \sqrt{3} )^2-10 \cdot 1=x^2 \cdot 1-\left(2 \cdot \frac{ \sqrt{3} }{2} + \sqrt{8} \cdot \frac{ \sqrt{2} }{2} \right)^2 \\ \\ (x+ \sqrt{3} )^2-10=x^2-\left(\sqrt{3} +2 \sqrt{2} \cdot \frac{ \sqrt{2} }{2} \right)^2 \\ \\ x^2+2 \cdot x \cdot \sqrt{3} + \sqrt{3}^2-10=x^2-( \sqrt{3} + \sqrt{2} \cdot \sqrt{2} )^2 \\ \\ x^2+2x \sqrt{3} +3-10=x^2-( \sqrt{3} +2)^2$
$\displaystyle \it x^2+2x \sqrt{3} +3-10=x^2-( \sqrt{3} ^2+2 \cdot \sqrt{3} \cdot 2+2^2) \\ \\ x^2+2x \sqrt{3} +3-10=x^2-(3+4 \sqrt{3} +4) \\ \\ x^2+2x \sqrt{3} +3-10=x^2-(7+4 \sqrt{3} ) \\ \\ x^2+2x \sqrt{3} +3-10=x^2-7-4 \sqrt{3} \\ \\ x^2-x^2+2x \sqrt{3} =-7-4 \sqrt{3} -3+10 \\ \\ 2x \sqrt{3} =-4 \sqrt{3} \Rightarrow x=- \frac{4 \sqrt{3} }{2 \sqrt{3} } \Rightarrow x=-2$