Question

Resolva as equações a seguir utilizando a fórmula de


bhaskara:


(a)x?+2x-10=0;


(b)x2-4x+2=0; (c)3x2–5x+2=0; (d)-x2


X-1=0 *

Answer 1

Resposta:

a) $x1 = -1 + \sqrt{11} \\\\x2 = -1 - \sqrt{11}$

b) $x1 = 2 + \sqrt{2} \\\\x2 = 2 - \sqrt{2}$

c) $x1 = \frac{5 + 1 }{6} = 1\\\\x2 = \frac{5 - 1}{6} = \frac{2}{3}$

d) $x = \frac{-1 +/- \sqrt{-3} }{-2}$

Explicação passo-a-passo:

a)

$x^{2} + 2x - 10 = 0\\\\delta = 2^{2} - (4 . 1 . -10)\\\\delta = 4 + 40\\\\delta = 44\\\\x = \frac{-2 +/- \sqrt{44} }{2. 1}\\\\x = \frac{-2 +/- 2\sqrt{11} }{2} \\\\x = -1 +/- \sqrt{11}$

b)

$x^{2} - 4x + 2 = 0\\\\delta = (-4)^{2} - (4 . 1 .2)\\\\delta = 16 - 8\\\\delta = 8\\\\x = \frac{-(-4) +/- \sqrt{8} }{2 . 1} \\\\x = \frac{4 +/- 2\sqrt{2} }{2} \\\\x = 2 +/- \sqrt{2}$

c)

$3x^{2} - 5x + 2 = 0\\\\delta = (-5)^{2} - (4 . 3 . 2)\\\\delta = 25 - 24\\\\delta = 1\\\\x = \frac{-(-5) +/- \sqrt{1} }{2 . 3}\\\\x = \frac{5 +/- 1}{6} \\\\x1 = \frac{5 + 1 }{6} = 1\\\\x2 = \frac{5 - 1}{6} = \frac{2}{3}$

d)

$-x^{2} + x - 1 =0\\\\delta = 1^{2} - (4 . -1 . -1)\\\\delta = 1 - 4\\\\delta = -3\\\\x = \frac{-1 +/- \sqrt{-3} }{2 . -1}\\\\x = \frac{-1 +/- \sqrt{-3} }{-2}$