Question

equação de segundo grau, ME AJUDEM PFV​

Answer 1

Resposta:

1)

a) $x^{2}$ - 49 = 0

$x^{2} = 49\\\\x = \sqrt{49}\\\\x = 7$

b)

$x^{2} = 1\\\\x = \sqrt{1}\\\\x = 1$

c)

$2x^{2} - 50 = 0\\\\2x^{2} = 50\\\\x^{2} = 25\\\\x = \sqrt{25}\\\\x = 5$

d)

$7x^{2} - 7 = 0\\\\7x^{2} = 7\\\\x^{2} = 1\\\\x = \sqrt{1}\\\\x = 1$

e)

$4x^{2} = 36\\\\x^{2} = 9\\\\x = \sqrt{9}\\\\x = 3$

f)

$5x^{2} - 15 = 0\\\\5x^{2} = 15\\\\x^{2} = 3\\\\x = \sqrt{3}$

g)

$21 = 7x^{2} \\\\x^{2} = 3\\\\x = \sqrt{3}$

h)

$5x^{2} + 20 = 0\\\\5x^{2} = -20\\\\x^{2} = -4\\\\x = \sqrt{-4}\\\\x = 2i$

i)

$4x^{2} - 49 = 0\\\\4x^{2} = 49\\\\x^{2} = \frac{49}{4}\\\\x = \sqrt{\frac{49}{4} } \\\\x = \frac{7}{2}$

j)

$16 = 9x^{2} \\\\x^{2} = \frac{16}{9}\\\\x = \sqrt{\frac{16}{9} }\\\\x = \frac{4}{3}$

L)

$3x^{2} + 30 = 0\\\\3x^{2} = -30\\\\x^{2} = -10\\\\x = \sqrt{-10}$

m)

$9x^{2} - 5 = 0\\\\9x^{2} = 5\\\\x^{2} = \frac{5}{9}\\\\x = \sqrt{\frac{5}{9} }\\\\x = \frac{\sqrt{5} }{3}$

2.

a)

$7x^{2} = 30 - 2\\\\7x^{2} = 28\\\\x^{2} = 4\\\\x = \sqrt{4}\\\\x = 2$

b)

$2x^{2} - 90 = 8\\\\2x^{2} = 98\\\\x^{2} = 49\\\\x = \sqrt{49}\\\\x = 7$

c)

$4x^{2} - 27 = x^{2} \\\\4x^{2} - x^{2} = 27\\\\3x^{2} = 27\\\\x^{2} = 9\\\\x = \sqrt{9}\\\\x = 3$

d)

$8x^{2} = 60 - 7x^{2} \\\\8x^{2} + 7x^{2} = 60\\\\15x^{2} = 60\\\\x^{2} = 4\\\\x = \sqrt{4}\\\\x = 2$

3.

a)

$3 (x^{2} - 1) = 24\\\\3x^{2} - 3 = 24\\\\3x^{2} = 27\\\\x^{2} = 9\\\\x = \sqrt{9}\\\\x = 3$

b)

$2 (x^{2} - 1) = x^{2} + 7\\\\2x^{2} - 2 = x^{2} + 7\\\\2x^{2} - x^{2} = 7 + 2\\\\x^{2} = 9\\\\x = \sqrt{9}\\\\ x = 3$

c)

$5 (x^{2} - 1) = 4 (x^{2} + 1)\\\\5x^{2} - 5 = 4x^{2} + 4\\\\5x^{2} - 4x^{2} = 4 + 5\\\\x^{2} = 9\\\\x = \sqrt{9}\\\\x = 3$

d)

$(x - 3) . (x + 4) + 8 = x\\\\x^{2} + 4x - 3x - 12 + 8 = x\\\\x^{2} + 0x - 4 = 0\\\\x^{2} = 4\\\\x = \sqrt{4}\\\\x = 2$