Question

Determine as raízes:
a) 2x² + X - 1 = 0
b) 2x² + 2x - 24 = 0
c) 3x² - 4x - 2 = -3
d) 1/4x² + 5x/4 - 6 = 0
e) X² + 5x - 2 = -8

Answer 1

$2x^2+x-1=0 \\ \\ Delta= 1+4.2.1= 9 \\ \\ \sqrt{Delta} = \sqrt{9} = 3 \\ \\ x= \frac{-1+3}{2.2} = \frac{2}{4} = \frac{1}{2} \\ \\ x_{1} = \frac{-1-3}{4} = \frac{-4}{4} = -1 \\ \\ 2 x^{2} +2x-24=0 \\ \\ Delta = 4+4.2.24= 196 \\ \\ \sqrt{Delta} = \sqrt{196} = 14 \\ \\ x= \frac{-2+14}{2.2} = \frac{12}{4} = 3 \\ \\ x_{1} = \frac{-2-14}{4} = \frac{-16}{4} = -4 \\ \\ 3 x^{2} -4x-2=-3 \\ \\ 3 x^{2} -4x-2+3=0 \\ \\ 3 x^{2} -4x+1=0 \\ \\ Delta = 16-4.3.1= 4$
$\sqrt{Delta} = \sqrt{4} = 2 \\ \\ x= \frac{4+2}{2.3} = \frac{6}{6} =1 \\ \\ x_{1} = \frac{4-2}{6} = \frac{2}{6} = \frac{1}{3} \\ \\ \frac{1}{4} x^{2} + \frac{5}{4} x-6 = 0 \\ \\ x^{2} +5x-6.4=0 \\ \\ x^{2} +5x-24=0 \\ \\ Delta= 25+4.1.24= 121 \\ \\ \sqrt{Delta} = \sqrt{121} = 11 \\ \\ x = \frac{-5+11}{2} = \frac{6}{2} = 3 \\ \\ x_{1} = \frac{-5-11}{2} = \frac{-16}{2} = -8 \\ \\ x^{2} +5x-2=-8 \\ \\ x^{2} +5x-2+8=0 \\ \\ x^{2} +5x+6=0 \\ \\ Delta=25-4.1.6=1$

$\sqrt{Delta} = \sqrt{1} = 1 \\ \\ x= \frac{-5+1}{2} = \frac{-4}{2} =-2 \\ \\ x_{1} = \frac{-5-1}{2} = \frac{-6}{2} = -3$