Question

Resolva as equações irracionais em R.

Answer 1

Explicação passo a passo:

$a)\sqrt{2x+2}=x+1\\\\ (\sqrt{2x+2})^2=(x+1)^2\\\\ 2x+2=(x+1)^2\\\\2x+2 = x^2+2x+1\\\\x^2+2x-1-2x-2=0\\\\x^2-3=0\\\\x^2=3\\\\x=+-\sqrt{3}$

$b)\sqrt{x+9}+x=11 \\\\\sqrt{x+9}=11-x \\\\(\sqrt{x+9})^2=(11-x)^2\\\\x+9=121-22x+x^2\\\\x^2-22x+121-x-9=0\\\\x^2-23x+112=0\\\\a=1 \\b= -23 \\c =112\\\\x= \frac{-b+-\sqrt{b^2-4.a.c} }{2.a} \\\\x= \frac{23+-\sqrt{23^2-4.1.112} }{2.1} \\\\x= \frac{23+-\sqrt{529-448} }{2} \\\\x= \frac{23+-\sqrt{81} }{2} \\\\x= \frac{23+-9 }{2} \\\\x_1 = \frac{23+9 }{2} \\\\x_1 = \frac{32 }{2} \\\\x_1 = 16\\\\x_2 = \frac{23-9 }{2} \\\\x_2= \frac{14 }{2} \\\\x_2 = 7$

$c) \sqrt{x+3}=x-3\\\\ (\sqrt{x+3})^2=(x-3)^2\\\\x+3 = x^2 -6x+9\\\\x^2-6x+9-x-3 = 0\\\\x^2-7x+6=0\\\\a=1\\b=-7\\c=6$$x= \frac{-b+-\sqrt{b^2-4.a.c} }{2.a} \\\\x= \frac{7+-\sqrt{7^2-4.1.6} }{2.1}$

$x= \frac{7+-\sqrt{49-24} }{2} \\\\x= \frac{7+-\sqrt{25} }{2} \\\\x= \frac{7+-5 }{2} \\\\x_1= \frac{7+5 }{2}\\\\x_1= \frac{12 }{2}\\\\x_1 = 6\\\\x_2= \frac{7-5 }{2}\\\\x_2= \frac{2 }{2}\\\\x_2 = 1$