Question

Determine as raízes ou zeros das funções abaixo, utilizando a fórmula de Bháskara.
a) X² - 4x – 5 = 0 Resposta x’= 5 e x”= -1
b) X² - x – 20 = 0 R. x’=5 e x”= -4
c) –x² + x + 12 = 0 R. X’ = -3 e X” = 4
d) X² - 3x – 4 = 0 R. x’ = 4 e x” = -1
e) X² - 8x + 7 = 0 R. x’ = 7 e x” = 1
f) 2x² - 7x = 0 R. X’ = 0 e x” = 7/2
g) 4x² - 16 = 0 R. x’ = 2 e x” = -2
h) X² - 5x + 6 = 0 R. X’ = 2 e x” = 3
i) X² - 5x + 8 = 0 R. x’ = 2 e x” = 2
j) –x² + 6x – 5 = 0 R. x’ = 1 e x” = 5
k) X² - 25 = 0 R. x’ = 5 e x” = -5
l) X² - 7x = 0 R. x’ = 0 e x” = 7

Answer 1

Resposta:

a) X² - 4x – 5 = 0 Resposta x’= 5 e x”= -1

a = 1

b = -4

c = -5

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-4) +- \sqrt{(-4)^{2} - 4 * 1 * -5} }{2*1}\\\\x = \frac{4+-\sqrt{16+20} }{2}\\\\x = \frac{4+-\sqrt{36} }{2}\\\\x = \frac{4+-6}{2}\\\\x' = \frac{4+6}{2}=\frac{10}{2}=5 \\\\x'' = \frac{4-6}{2}=\frac{-2}{2} =-1$

b) X² - x – 20 = 0 R. x’=5 e x”= -4

a = 1

b = -1

c = -20

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-1) +- \sqrt{(-1)^{2} - 4 * 1 * -20} }{2*1}\\\\x = \frac{1+-\sqrt{1+80} }{2}\\\\x = \frac{1+-\sqrt{81} }{2}\\\\x = \frac{1+-9}{2}\\\\x' = \frac{1+9}{2}=\frac{10}{2} =5\\\\x'' = \frac{1-9}{2}=\frac{-8}{2} =-4$

c) –x² + x + 12 = 0 R. X’ = -3 e X” = 4

a = -1

b = 1

c = 12

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(1) +- \sqrt{(1)^{2} - 4 * -1 * 12} }{2*-1}\\\\x = \frac{-1+-\sqrt{1+48} }{-2}\\\\x = \frac{-1+-\sqrt{49} }{-2}\\\\x = \frac{-1+-7}{-2}\\\\x' = \frac{-1+7}{-2}=\frac{6}{-2}= -3\\\\x'' = \frac{-1-7}{-2}=\frac{-8}{-2} =4$

d) X² - 3x – 4 = 0 R. x’ = 4 e x” = -1

a = 1

b = -3

c = -4

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-3) +- \sqrt{(-3)^{2} - 4 * 1 * -4} }{2*1}\\\\x = \frac{3+-\sqrt{9+16} }{2}\\\\x = \frac{3+-\sqrt{25} }{2}\\\\x = \frac{3+-5}{2}\\\\x' = \frac{3+5}{2}=\frac{8}{2}=4 \\\\x'' = \frac{3-5}{2}=\frac{-2}{2} =-1$

e) X² - 8x + 7 = 0 R. x’ = 7 e x” = 1

a = 1

b = -8

c = 7

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-8) +- \sqrt{(-8)^{2} - 4 * 1 * 7} }{2*1}\\\\x = \frac{8+-\sqrt{64-28} }{2}\\\\x = \frac{8+-\sqrt{36} }{2}\\\\x = \frac{8+-6}{2}\\\\x' = \frac{8+6}{2} =\frac{14}{2} =7\\\\x'' = \frac{8-6}{2}=\frac{2}{2} =1$

f) 2x² - 7x = 0 R. X’ = 0 e x” = 7/2

a = 2

b = -7

c = 0

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-7) +- \sqrt{(-7)^{2} - 4 * 2 * 0} }{2*2}\\\\x = \frac{7+-\sqrt{49+0} }{4}\\\\x = \frac{7+-\sqrt{49} }{4}\\\\x = \frac{7+-7}{4}\\\\x' = \frac{7+7}{4}=\frac{14}{4}=\frac{7}{2} \\\\x'' = \frac{7-7}{4}=\frac{0}{4} =0$

g) 4x² - 16 = 0 R. x’ = 2 e x” = -2

a = 4

b = 0

c = -16

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(0) +- \sqrt{(0)^{2} - 4 * 4 * -16} }{2*4}\\\\x = \frac{0+-\sqrt{0+256} }{8}\\\\x = \frac{0+-\sqrt{256} }{8}\\\\x = \frac{0+-16}{8}\\\\x' = \frac{0+16}{8}=\frac{16}{8}=2 \\\\x'' = \frac{0-16}{8}=\frac{-16}{8}=-2$

h) X² - 5x + 6 = 0 R. X’ = 2 e x” = 3

a = 1

b = -5

c = 6

$\lim_{n \to \infty} a_n x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-5) +- \sqrt{(-5)^{2} - 4 * 1 * 6} }{2*1}\\\\x = \frac{5+-\sqrt{25-24} }{2}\\\\x = \frac{5+-\sqrt{1} }{2}\\\\x = \frac{5+-1}{2}\\\\x' = \frac{5+1}{2}=\frac{6}{2}=3 \\\\x'' = \frac{5-1}{2}=\frac{4}{2} =2$

i) X² - 5x + 8 = 0 R. x’ = 2 e x” = 2

a = 1

b = -5

c = 8

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-5) +- \sqrt{(-5)^{2} - 4 * 1 * 8} }{2*1}\\\\x = \frac{5+-\sqrt{25-32} }{2}\\\\x = \frac{5+-\sqrt{-7} }{2}\\\\x' = \frac{5+\sqrt{-7} }{2}\\\\x'' = \frac{5-\sqrt{-7} }{2}$

para este resultado ser verdadeiro x’ = 2 e x” = 2, somente se b=-4 e c=4

j) –x² + 6x – 5 = 0 R. x’ = 1 e x” = 5

a = -1

b = 6

c = -5

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(6) +- \sqrt{(6)^{2} - 4 * -1 * -5} }{2*-1}\\\\x = \frac{-6+-\sqrt{36-20} }{-2}\\\\x = \frac{-6+-\sqrt{16} }{-2}\\\\x = \frac{-6+-4}{-2}\\\\x' = \frac{-6+4}{-2}=\frac{-2}{-2}=1 \\\\x'' = \frac{-6-4}{-2}=\frac{-10}{-2}=5$

k) X² - 25 = 0 R. x’ = 5 e x” = -5

a = 1

b = 0

c = -25

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(0) +- \sqrt{(0)^{2} - 4 * 1 * -25} }{2*1}\\\\x = \frac{0+-\sqrt{0+100} }{2}\\\\x = \frac{0+-\sqrt{100} }{2}\\\\x = \frac{0+-10}{2}\\\\x' = \frac{0+10}{2}=\frac{10}{2}=5 \\\\x'' = \frac{0-10}{2}=\frac{-10}{2}=-5$

l) X² - 7x = 0 R. x’ = 0 e x” = 7

a = 1

b = -7

c = 0

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-7) +- \sqrt{(-7)^{2} - 4 * 1 * 0} }{2*1}\\\\x = \frac{7+-\sqrt{49+0} }{2}\\\\x = \frac{7+-\sqrt{49} }{2}\\\\x = \frac{7+-7}{2}\\\\x' = \frac{7+7}{2}=\frac{14}{2} =7\\\\x'' = \frac{7-7}{2}=\frac{0}{2} =0$

espero ter ajudado.