Question

me ajudem amanha preciso entregar isso e ainda n fiz

Answer 1

b) $3x^2=16-2x-4 \\3x^2=12-2x \\0=12-2x-3x^2 \\ \\a=-3 \\b=-2 \\c=12 \\ \\X= \frac{-b+/- \sqrt{b^2-4ac} }{2a} \\ \\X= \frac{2+/- \sqrt{-2^2-4*-3*12} }{2*-3} \\ \\X= \frac{2+/- \sqrt{148} }{-6}$ 

c) $x^2=7x+10 \\0=7x+10-x^2 \\ \\a=-1 \\b=7 \\c=10 \\ \\ \frac{-7+/- \sqrt{7^2-4*-1*10} }{2*-1} \\ \\ \frac{-7+/- \sqrt{7^2-4*-1*10} }{2*-1} \\ \\ \frac{-7+/- \sqrt{89} }{-2}$

d) $x^2+2(x+1)=5 \\x^2+2x+2=5 \\x^2+2x+2-5=0 \\x^2+2x-3=0 \\ \\a=1 \\b=2 \\c=-3 \\ \\ \frac{-2+/- \sqrt{2^2-4*1*-3} }{2*1} \\ \\ \frac{-2+/- \sqrt{16} }{2} \\ \\ \frac{-2+/- 4 }{2} \\ \\x'=\frac{-2+ 4 }{2} = 2/2 = 1 \\ \\x''=\frac{-2- 4 }{2} = -6/2 = -3$

e) $2x(4x-2)-4=0 \\8x^2-4x-4=0 \\ \\a=8 \\b=-4 \\c=-4 \\ \\ \frac{4+/- \sqrt{-4^2-4*8*-4}}{2*8} \\ \\ \frac{4+/- \sqrt{16-4*8*-4}}{16} \\ \\ \frac{4+/- \sqrt{144}}{16} \\ \\ \frac{4+/- 12}{16} \\ \\x'= \frac{4+12}{16} = 16/16 = 1 \\ \\x''=\frac{4- 12}{16} =-8/16=-0,5$

f) $x^2+(x+1)^2=25 \\x^2+(x+1)(x+1)=25 \\x^2+x^2+2x+1=25 \\x^2+x^2+2x+1-25=0 \\2x^2+2x-24=0 \\ \\a=2 \\b=2 \\c=-24 \\ \\ \frac{-2+/- \sqrt{2^2-4*2*-24} }{2*2} \\ \\ \frac{-2+/- \sqrt{196} }{4} \\ \\ \frac{-2+/-14 }{4} \\ \\x'= \frac{-2+14 }{4} = 12/4=3 \\ \\x''= \frac{-2-14 }{4} = -16/4=-4$

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