Question

1) Aplique Bhaskara nas equações do 2°grau a seguir,
e determine as raízes (caso existam) das seguintes
equações do 2º grau
a) x2 + 2x -3 = 0
b) -x2 + 2x +2 = 0
c) x2 + 4x -5 = 0
d) - 2x2 + 2 = 0​

Answer 1

Explicação passo-a-passo:

.

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

$a = 1$

$b = 2$

$c = - 3$

$∆ = b {}^{2} - 4ac$

$∆ = 2 {}^{2} - 4 \times 1 \times - 3$

$∆ = 4 + 12$

$∆ = 16$

$\dfrac{ - b \dfrac{ + }{} \sqrt{∆} }{2a}$

$\dfrac{ - 2 \dfrac{ + }{} \sqrt{16} }{2 \times 1}$

$\dfrac{ - 2 \dfrac{ + }{} 4}{2}$

$x {}^{1} = \dfrac{ - 2 + 4}{2} = \dfrac{2}{2} = 1$

$x {}^{2} = \dfrac{ - 2 - 4}{2} = \dfrac{ - 6}{2} = - 3$

.

$- x {}^{2} + 2x + 2 = 0$

$x {}^{2} - 2x - 2 = 0$

$x = \dfrac{ - ( - 2) \frac{ + }{} \sqrt{( - 2) {}^{2} - 4 \times 1 \times - 2 } }{2 \times 1}$

$x = \dfrac{ 2 \frac{ + }{} \sqrt{4 + 8} }{2}$

$x = \dfrac{2 \frac{ + }{} \sqrt{12} }{2}$

$x = \dfrac{2 \frac{ + }{}2 \sqrt{3} }{2}$

$x = \dfrac{2 + 2 \sqrt{3} }{2}$

$x = \dfrac{2 - 2 \sqrt{3} }{2}$

$x {}^{1} = 1 + \sqrt{3}$

$x {}^{2} = 1 - \sqrt{3}$

.

$x {}^{2} + 4x - 5 = 0$

$a = 1$

$b = 4$

$c = - 5$

$∆ = b {}^{2} - 4ac$

$∆ = 4 {}^{2} - 4 \times 1 \times - 5$

$∆ = 16 + 20$

$∆ = 36$

$\dfrac{ - b \dfrac{ + }{} \sqrt{∆} }{2a}$

$\dfrac{ - 4 \dfrac{ + }{} \sqrt{36} }{2 \times 1}$

$\dfrac{ - 4 \dfrac{ + }{}6 }{2}$

$x {}^{1} = \dfrac{ - 4 + 6}{2} = \dfrac{2}{2} = 1$

$x {}^{2} = \dfrac{ - 4 - 6}{2} = - \dfrac{10}{2} = - 5$

.

$- 2x {}^{2} + 2 = 0$

$- 2x {}^{2} = 0 - 2$

$- 2x {}^{2} = - 2( - 1)$

$2x {}^{2} = 2$

$x {}^{2} = \dfrac{2}{2}$

$x {}^{2} = 1$

$x = \frac{ + }{} \sqrt{1}$

$x = ±1$

$s = ( + 1 \: e \: - 1)$