Question

x elevado a 4 - 8x² - 9 = 0

x elevado a 4 - 8 x² + 16=0

(x²-1) (x²-12) +24 =0

(x²+2)² = 2 . (x² + 6) = 0

Answer 1

$x^{4} -8 x^{2} +16=0\to x^{2} =y\\\\ y^{2} -8y+16=0\\\\ a=1;b=-8;c=16\\\\\\ \Delta=b^2-ac\to \Delta=(-8)^2-4*1*16\to \Delta=64-64\to \Delta=0\\\\\\ y= \frac{-b+- \sqrt{\Delta} }{2a} \to y= \frac{-(-8)+- \sqrt{0} }{2*1} \to y= \frac{8+- 0}{2}\to \\\\\\ y'=y''= \frac{8}{2} \to y'=y''4\\\\\\ S=(4)$

Obs: Essa equação tem apenas uma solução, pois, $\Delta=0$.






$(x^2-1) (x^2-12) +24 =0\\\\ x^{4} -12 x^{2} - x^{2} +12+24=0\\\\ x^{4} -13 x^{2} +36=0\to x^{2} =y\\\\\\ y^{2} -13y+36=0\\\\ a=1;b=-13;c=36\\\\\\ \Delta=b^2-4ac\to \Delta=(-13)^2-4*1*36\to \Delta=169-144\to \Delta=25\\\\\\ y= \frac{-b+- \sqrt{\Delta} }{2a} \to y= \frac{-(-13)+- \sqrt{25} }{2*1} \to y= \frac{13+- 5 }{2} \to \\\\\\ y'= \frac{13+ 5 }{2} \to y'= \frac{18}{2} \to y'=9\\\\\\ y''= \frac{13- 5 }{2} \to y''= \frac{8 }{2} \to y''=4\\\\\\ S=(4;9)$






$(x^{2} +2)^2 = 2 . ( x^{2} + 6) \\\\ x^{4} +2 x^{2} +2 x^{2} +4=2 x^{2} +12\\ x^{4} +2 x^{2} +2 x^{2} +4-2 x^{2} -12=0\\\\ x^{4} + 2^{2} -8=0\to x^{2} =y\\\\ y^{2} +2y-8=0\\\\ a=1;b=2;c=-8\\\\\\ \Delta=b^2-4ac\to \Delta=2^2-4*1*(-8)\to \Delta=169-1444+32 \\ \Delta=36\\ \\\\ y= \frac{-b+- \sqrt{\Delta} }{2a} \to y= \frac{-2+- \sqrt{36} }{2*1} \to y= \frac{-2+- 6 }{2}\to \\\\\\ y'= \frac{-2+6 }{2} \to y'= \frac{4 }{2} \to y'=2\\ \\\\ y''= \frac{-2- 6 }{2} \to y''=\frac{-8 }{2} \to y''=-4\\\\\\ S=(-4;2)$





Pronto Bárbara, ufa, essa foi longa, rss... Espero que tenha ajudado...