Question

determine as soluções reais das equações
a)  2$x^{2}$ - 3x + 1 =0
b) $x^{2}$ -2x - 3 =0

Answer 1

a)

$2x^{2}-3x+1=0$

$\Delta=b^{2}-4ac\\\Delta=(-3)^{2}-4*2*1\\\Delta=9-8\\\Delta=1$

$x=(-b\pm\sqrt{\Delta})/2a\\x=(-[-3]\pm\sqrt{1})/(2*2)\\x=(3\pm1)/4\\\\x'=(3+1)/4\\x'=4/4=\\x'=1\\\\x''=(3-1)/4\\x''=2/4\\x''=1/2$

$\boxed{\boxed{S=\{\frac{1}{2},~1\}}}$

b)

$x^{2}-2x-3=0\\\\\Delta=b^{2}-4ac\\\Delta=(-2)^{2}-4*1*(-3)\\\Delta=4+12\\\Delta=16$

$x=(-b\pm\sqrt{\Delta})/2a\\x=(-[-2]\pm\sqrt{16})/(2*1)\\x=(2\pm4)/2\\\\x'=(2+4)/2\\x'=6/2\\x'=3\\\\x''=(2-4)/2\\x''=-2/2\\x''=-1\\\\\boxed{\boxed{S\{-1,3\}}}$