Question

Resolva as funções: A) 25 x elevado a 2 = 20 × -4 B) 2x= 15 - × elevado a 2 C)x elevado a 2 + 3 × - 6 = -8 D) × elevado a 2 + × - 7 = 5 E) 4 × elevado a 2 - x + 1 = × + 3 × elevado a 2

Answer 1

Olá Fernanda

A)
$25x^{2} = 20\times (-4)\\ 25x^{2} = -80$
x∉R
Quando o membro esquerdo é ≤ 0 ,não há solução


B) 
$2x= 15 - x^{2}\\ x^{2}+2x - 15=0\\ \\ x={ \frac{-b \frac{+}{} \sqrt{b^{2}-4\times a\times c} }{2\times a} }\\\\ x={ \frac{-2 \frac{+}{} \sqrt{2^{2}-4\times 1\times (-15)} }{2\times 1} }\\\\ x={ \frac{-2 \frac{+}{} \sqrt{4+60} }{2} }\\\\ x={ \frac{-2 \frac{+}{} \sqrt{64} }{2} }\\\\ x={ \frac{-2 \frac{+}{} 8 }{2} }\\\\ x'={ \frac{-2 + 8 }{2} } ={ \frac{6}{2} } = 3 \\\\ x"={ \frac{-2 - 8 }{2} } = { \frac{-10}{2} } =-5$


C)
$x^{2} + 3 x - 6 = -8\\ x^{2}+3x+2=0\\\\ x={ \frac{-b \frac{+}{} \sqrt{b^{2}-4\times a\times c} }{2\times a} }\\\\ x={ \frac{-3 \frac{+}{} \sqrt{3^{2}-4\times 1\times 2} }{2\times 1} }\\\\ x={ \frac{-3 \frac{+}{} \sqrt{9-8} }{2} }\\\\ x={ \frac{-3 \frac{+}{} \sqrt{1} }{2} }\\\\ x={ \frac{-3 \frac{+}{} 1 }{2} }\\\\ x'={ \frac{-3 + 1 }{2} } ={ \frac{-2}{2} } = -1 \\\\ x"={ \frac{-3 - 1 }{2} } ={ \frac{-4}{2} } =-2$


D)
$x^{2}+x-7=5\\ x^{2}+x-12=0\\\\ x={ \frac{-b \frac{+}{} \sqrt{b^{2}-4\times a\times c} }{2\times a} }\\\\ x={ \frac{-1 \frac{+}{} \sqrt{1^{2}-4\times 1\times -12} }{2\times 1} }\\\\ x={ \frac{-1 \frac{+}{} \sqrt{1+48} }{2} }\\\\ x={ \frac{-1 \frac{+}{} \sqrt{49} }{2} }\\\\ x={ \frac{-1 \frac{+}{} 7}{2} }\\\\ x'={ \frac{-1 + 7 }{2} } ={ \frac{6}{2} } = 3 \\\\ x"={ \frac{-1 - 7 }{2} } ={ \frac{-8}{2} } =-4$


E)
$4x^{2}-x+1=x+3x^{2}\\ x^{2}-2x+1=0\\ (x-1)^{2}=0\\ x-1=0\\ x=1$


Espero ter ajudado
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