Question

Como resolve essas equações?

A) (x-4)-(x-1)²=10-x
B)(x+1) (2x+3)=4x²-22​

Answer 1

Resposta:

$A) \frac{4+-\sqrt{-44} }{2} \\B)x'=\frac{5}{-2} \\x''=5$

Explicação:

A)

$(x-4)-(x-1)^{2} =10-x\\x-4(x^{2} -2x+1)=10-x\\x-4-x^{2} +2x-1=10-x\\x-4-x^{2} +2x-1=10-x\\3x-4-x^{2} -1=10-x\\3x-5-x^{2} =10-x\\4x-5-x^{2} -10=0\\4x-15-x^{2} =0$

$(-x^{2} +4x-15)*(-1)=0\\x^{2}-4x+15=0 \\y=\frac{-(b)+-\sqrt{b^{2}-4ac } }{2a}\\y=\frac{4+-\sqrt{16-60} }{2}\\ y=\frac{4+-\sqrt{-44} }{2}$

B)

$(x+1)(2x+3)=4x^{2} -22\\2x^{2} +3x+2x+3=4x^{2} -22\\-2x^{2} +5x+25\\y=\frac{-(b)+-\sqrt{b^{2}-4ac } }{2a} \\y = \frac{-5+-\sqrt{25+200} }{-4}\\y = \frac{-5+-\sqrt{225} }{-4}  \\y=\frac{-5+-15 }{-4}\\y'=\frac{-5+15}{-4} \\y'=\frac{10}{-4}  \\y'=\frac{5}{-2}\\y''=\frac{-5-15}{-4}\\y''=\frac{-20}{-4}\\y''=5$

Espero ter ajudado

;)