Question

de o conjunto solução das equações a seguir: (x²+2x-8)(-x²-2x+15)>0 e 2x²-8/-x²+3x sendo maior ou igual a zero

Answer 1

Resposta:

 x2−2x−8=0a=1;b=−2;c=−8Δ=b2−4ac→Δ=(−2)2−4∗1∗(−8)→Δ=4+32→Δ=36x=2a−b+−Δ→x=2∗1−(−2)+−36→x=22+−6→x′=22+6→x′=28→x′=4x′′=22−6→x′′=2−4→x′′=−2S=(−2;4)

\begin{gathered}b) x^{2} -3x+4=6-2x\\\\\\ x^{2} -3x+4-6+2x=0\to\\\\ x^{2} -x-2=0\\\\ a=1;b=-1;c=-2\\\\ \Delta=b^2-4ac\to \Delta=(-1)^2-4*1*(-2)\to \Delta=1+8\to \Delta=9\\\\ x= \frac{-b+- \sqrt{\Delta} }{2a} \to x= \frac{-(-1)+- \sqrt{9} }{2*1} \to x= \frac{1+- 9 }{2} \to \\\\ x'= \frac{1+9 }{2} \to x'= \frac{10 }{2} \to x'=5\\\\ x''= \frac{1- 9 }{2} \to x''= \frac{-8 }{2} \to x''=-4\\\\\\ S=(-4;5)\end{gathered}b)x2−3x+4=6−2xx2−3x+4−6+2x=0→x2−x−2=0a=1;b=−1;c=−2Δ=b2−4ac→Δ=(−1)2−4∗1∗(−2)→Δ=1+8→Δ=9x=2a−b+−Δ→x=2∗1−(−1)+−9→x=21+−9→x′=21+9→x′=210→x′=5x′′=21−9→x′′=2−8→x′′=−4S=(−4;5)

Acho que é isso