Question

escreva as equações abaixo na forma geral e resolva em R​

Answer 1

Explicação passo-a-passo:

A)

$2x {}^{2} - 7x = 15$

$2x {}^{2} - 7x - 15 = 0$

$2x {}^{2} + 3x - 10x - 15 = 0$

$x(2x + 3) - 5(2x + 3) = 0$

$(2x + 3)(x - 5)$

$2x + 3 = 0$

$2x = 0 - 3$

$2x = - 3$

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

$x - 5 = 0$

$x = 0 + 5$

$x {}^{2} = 5$

B)

$4x {}^{2} + 9 = 12x$

$4x {}^{2} + 9 - 12x = 0$

$4x {}^{2} - 12x + 9 = 0$

$(2x - 3) {}^{2} = 0$

$2x - 3 = 0$

$2x = 0 + 3$

$2x = 3$

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

C)

$x {}^{2} = x + 12$

$x {}^{2} - x - 12 = 0$

$x {}^{2} + 3x - 4x - 12 = 0$

$x(x + 3) - 4(x + 3) = 0$

$x + 3 = 0$

$x = 0 - 3$

$x {}^{1} = - 3$

$x - 4 = 0$

$x = 0 + 4$

$x {}^{2} = 4$

D)

$2x {}^{2} = - 12x - 18$

$2x {}^{2} = - 12x - 18( \div 2)$

$x {}^{2} = - 6x + 9$

$x {}^{2} + 6x + 9 = 0$

$(x + 3) {}^{2} = 0$

$x + 3 = 0$

$x = 0 - 3$

$x = - 3$

E)

$x {}^{2} + 9 = 4x$

$x {}^{2} + 9 - 4x = 0$

$x {}^{2} - 4x + 9 = 0$

$x = \dfrac{ - ( - 4) \frac{ + }{} \sqrt{( - 4) {}^{2} - 4 \times 1 \times 3} }{2 \times 1}$

$x = \dfrac{4 \frac{ + }{} \sqrt{16 - 36} }{2}$

$x = \dfrac{4 \frac{ + }{} \sqrt{ - 20} }{2}$

Não possui solução no conjunto dos números reais.

F)

$25x {}^{2} = 20x - 4$

$25x {}^{2} - 20x + 4 = 0$

$(5x - 2) {}^{2} = 0$

$5x - 2 = 0$

$5x = 0 + 2$

$5x = 2$

$x = \dfrac{2}{5}$

G)

$2x {}^{2} = - 5 - 7x$

$2x {}^{2} + 5 + 7x = 0$

$2x {}^{2} + 7x + 5 = 0$

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

$x(2x + 5) + 2x + 5 = 0$

$(2x + 5)(x + 1) = 0$

$2x + 5 = 0$

$2x = 0 - 5$

$2x = - 5$

$x {}^{1} = - \dfrac{5}{2}$

$x + 1 = 0$

$x = 0 - 1$

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

H)

$2x {}^{2} = - 3 + 7x$

$2x {}^{2} + 3 - 7x = 0$

$2x {}^{2} - 7x + 3 = 0$

$2x {}^{2} - x - 6x + 3 = 0$

$x(2x - 1) - 3(2x - 1) = 0$

$(2x - 1)(x - 1) = 0$

$2x - 1 = 0$

$2x = 0 + 1$

$2x = 1$

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

$x - 3 = 0$

$x = 0 + 3$

$x {}^{2} = 3$

I)

$2x = 15 - x {}^{2}$

$2x - 15 + x {}^{2} = 0$

$x {}^{2} + 2x - 15 = 0$

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

$x(x + 5) - 3(x + 5) = 0$

$(x + 5)(x - 3) = 0$

$x + 5 = 0$

$x = 0 - 5$

$x {}^{1} = - 5$

$x - 3 = 0$

$x = 0 + 3$

$x {}^{2} = 3$

J)

$7x - 12 = x {}^{2}$

$7x - 12 - x {}^{2} = 0$

$- 7x + 12 + x {}^{2} = 0$

$x {}^{2} - 7x + 12 = 0$

$x {}^{2} - 3x - 4x + 12 = 0$

$x(x - 3) - 4(x - 3) = 0$

$(x - 3)(x - 4) = 0$

$x - 3 = 0$

$x = 0 + 3$

$x {}^{1} = 3$

$x - 4 = 0$

$x = 0 + 4$

$x {}^{2} = 4$