dashKoKOkko
dashKoKOkko
03-06-2017
Mathematics
contestada
(5-x)(5+x)=(x-4)²-7
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Tucon
Tucon
03-06-2017
[tex]\displaystyle\\ (5-x)(5+x)=(x-4)^2-7\\\\ 5^2 - x^2 = x^2-8x + 16 -7\\\\ -x^2 -x^2+8x + 25 -16+7=0\\\\ -2x^2 + 8x +16=0\\\\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{-8\pm \sqrt{64-4\cdot(-2)\cdot 16} }{2\cdot(-2)}=\\\\ = \frac{-8\pm \sqrt{64+ 128} }{-4}= \frac{-8\pm \sqrt{192} }{-4}= \frac{-8\pm \sqrt{64\cdot 3} }{-4}=\\\\ =\frac{-8\pm 8\sqrt{ 3} }{-4}=2\mp 2\sqrt{ 3}\\\\ \boxed{\bf x_1 = 2-2\sqrt{ 3}}\\\\ \boxed{\bf x_2 = 2+2\sqrt{ 3}}[/tex]
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