Question

Resolver la siguiente ecuación:

a) (t-4)^{2}=(t+4)^{2}+32

b) \sqrt{3x} +\sqrt{12} =\frac{x+5}{\sqrt{3} }

Answer 1

a)

$(t-4)^{2} =(t+4)^{2}+32\\\\(t-4)(t-4)=(t+4) (t+4) +32\\\\t^{2} -8t+16=t^{2} +8t+16+32\\\\-8t+16=8t+48\\\\16-48=8t+8t\\\\-32=16t\\\\t=-\frac{32}{16} \\\\t=-2$

b)

$\sqrt{3x} +\sqrt{12} =\frac{x+5}{\sqrt{3} }\\\\\sqrt{3}( \sqrt{3x} +\sqrt{12}) =x+5\\\\\sqrt{3} \sqrt{3x} +\sqrt{3} \sqrt{12}=x+5\\\\\sqrt{3}\sqrt{3x}+\sqrt{36}=x+5\\\\\sqrt{3}\sqrt{3x}+6=x+5\\\\\sqrt{3}\sqrt{3x}=x+5-6\\\\\sqrt{3}\sqrt{3x}=x-1\\\\(\sqrt{3}\sqrt{3x})^{2} =(x-1)^{2} \\\\9x=x^{2} -2x+1\\\\x^{2} -11x+1=0\\\\x=\frac{11+-\sqrt{121-4} }{2}$

$x=\frac{11+-\sqrt{117} }{2} =\frac{11+\sqrt{117} }{2}$