Question

2) Determine as raízes (zeros) reais de cada uma das funções de R em R
dadas pelas seguintes leis:
a) y= x2 + x -3
b) y = 2x2 – 8x -10
c) y = x2 – 2x – 80
d) y = x2- 3x – 10
e) y = x2 + 3x - 40
AJUDA PFV!!!

Answer 1

2)

a)

${x}^{2} + x - 3 = 0\\ ∆ = {1}^{2} - 4 \times 1 \times - 3 = 1 + 12 = 13 \\ x1 = \frac{ - 1 + \sqrt{13} }{2} \\ x2 = \frac{ - 1 - \sqrt{13} }{2}$

b)

$2 {x}^{2} - 8x - 10 = 0 \\ ∆ = ( - 8) {}^{2} - 4 \times 2 \times - 10 = 64 + 80 = 144 \\ x1 = \frac{ - ( - 8) + 12}{4} = \frac{8 + 12}{4} = \frac{20}{4} = 5 \\ x2 = \frac{8 - 12}{4} = \frac{ - 4}{4} = - 1$

c)

${x}^{2} - 2x - 80 = 0\\ ∆ = ( - 2) {}^{2} - 4 \times 1 \times - 80 = 4 + 320 = 324 \\ x1 = \frac{ - ( - 2) + 18}{2} = \frac{2 + 18}{2} = \frac{20}{2} = 10 \\ x2 = \frac{2 - 18}{2} = \frac{ - 16}{2} = - 8$

d)

${x}^{2} - 3x - 10 = 0 \\ ∆ = ( - 3) {}^{2} - 4 \times 1 \times - 10 = 9 + 40 = 49 \\ x1 = \frac{ - ( - 3) + 7}{2} = \frac{3 + 7}{2} = \frac{10}{2} = 5 \\ x2 = \frac{3 - 7}{2} = \frac{ - 4}{2} = - 2$

e)

${x}^{2} + 3x - 40 = 0 \\ ∆ = {3}^{2} - 4 \times 1 \times - 40 = 9 + 160 = 169 \\ x1 = \frac{3 + 13}{2} = \frac{16}{2} = 8 \\ x2 = \frac{3 - 13}{2} = \frac{ - 10}{2} = - 5$