Question

Resolva, em R, as equações seguintes:
a) |3x - 2| = 1 d) |x² - 4| = 5
b) |x + 6 |= 4 e) ||2x - 1| - 3| = 2
c) |x² - 2x - 5|=3
​

Answer 1

a)

$|3x - 2| = 1 \\ 3x - 2 = 1 \\ 3x = 2 + 1 \\ 3x = 3$

$x = \frac{3}{3} \\ x = 1$

$3x - 2 = - 1 \\ 3x = 2 - 1 \\ 3x = 1 \\ x = \frac{1}{3}$

S={1,⅓}

b)

$|x + 6| = 4 \\ x + 6 = 4 \\ x = 4 - 6 \\ x = - 2$

$x + 6 = - 4 \\ x = - 4 - 6 \\ x = - 10$

S={-10,-2}

c)

$| {x}^{2} - 2x - 5 | = 3 \\ {x}^{2} - 2x - 5 = 3 \\ {x}^{2} - 2x - 5 - 3 = 0 \\ {x}^{2} - 2x - 8 = 0$

$\Delta = 4 + 32 = 36 \\ x = \frac{2±6}{2} \\ x' = \frac{2 + 6}{2} = 4 \\ x'' = - 2$

${x}^{2} - 2x - 5 = - 3 \\ {x}^{2} - 2x - 5 + 3 = 0 \\ {x}^{2} -2x - 2 = 0$

$\Delta = 4 + 8 = 12 \\ x = \frac{ - 2±2 \sqrt{3} }{2} \\ x = - 1± \sqrt{3} </p><p>$

$x' = - 1 + \sqrt{3} \\ x'' = - 1 - \sqrt{3}$

S={-1-√3,-2,-1+√3,4}

d)

$| {x}^{2} - 4| = 5 \\ {x}^{2} - 4 = 5 \\ {x}^{2} = 5 + 4 \\ {x}^{2} = 9 \\ x = ± \sqrt{9} = ±3$

${x}^{2} - 4 = - 5 \\ {x}^{2} = 4 - 5 \\ {x}^{2} = - 1$

S={-3,3}

e)

$||2x - 1| - 3| = 2 \\ |2x - 1| - 3 = 2 \\ |2x - 1| = 2 + 3 \\ |2x - 1| = 5$

$2x - 1 = 5 \\ 2x = 5 + 1 \\ 2x = 6 \\ x = \frac{6}{2} \\ x = 3$

$2x - 1 = - 5 \\ 2x = 1 - 5 \\ 2x = - 4 \\ x = - \frac{4}{2} \\ x = - 2$

$|2x - 1| - 3 = - 2 \\ |2x - 1| = 3 - 2 \\ |2x - 1| = 1$

$2x - 1 = 1 \\2x = 1 + 1 \\ 2x = 2 \\ x = \frac{2}{2} \\ x = 1$

$2x - 1 = - 1 \\ 2x = 1 - 1 \\ 2x = 0 \\ x = \frac{0}{2} \\ x = 0$

S={-2,0,1,3}