Question

Escreva as equações abaixo na forma geral e resolva em R:​

Answer 1

Explicação passo-a-passo:

A)

$x(x + 3) - 40 = 0$

$x {}^{2} + 3x - 40 = 0$

$x = \dfrac{ - 3 \frac{ + }{} \sqrt{3 {}^{2} \times 4 \times 1 \times - 40 } }{2}$

$x = \dfrac{ - 3 \frac{ + }{} \sqrt{9 + 160} }{2}$

$x = \dfrac{ - 3 \frac{ + }{} \sqrt{169} }{2}$

$x = \dfrac{ - 3 \frac{ + }{}13 }{2}$

$x {}^{1} = \dfrac{ - 3 + 13}{2} = \dfrac{10}{2} = 5$

$x {}^{2} = \dfrac{ - 3 - 13}{2} = - \dfrac{16}{2} = - 8$

B)

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

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

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

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

$x = \dfrac{ - ( - 1) \frac{ + }{} \sqrt{( - 1) {}^{2} - 4 \times 1 \times - 5 } }{2 \times 1}$

$x = \dfrac{1 \frac{ + }{} \sqrt{1 + 20} }{2}$

$x = \dfrac{1 \frac{ + }{} \sqrt{21} }{2}$

$x {}^{1} = \dfrac{1 + \sqrt{21} }{2}$

$x {}^{2} = \dfrac{1 - \sqrt{21} }{2}$

C)

$10 + x(x - 2) = 2$

$10 + x {}^{2} - 2x = 2$

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

$8 + x {}^{2} - 2x = 0$

$x {}^{2} - 2x + 8 = 0$

$x = \dfrac{ - ( - 2) \frac{ + }{} \sqrt{( - 2) {}^{2} - 4 \times 1 \times 8 } }{2 \times 1}$

$x = \dfrac{2 \frac{ + }{} \sqrt{4 - 32} }{2}$

$x = \dfrac{2 \frac{ + }{} \sqrt{ - 28} }{2}$

Não possui solução no conjunto dos números reais.

D)

$(x - 1)(x + 5) = 7$

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

$x {}^{2} + 4x - 5 = 7$

$x {}^{2} + 4x - 5 - 7 = 0$

$x {}^{2} + 6x - 2x - 12 = 0$

$x(x + 6) - 2(x + 6) = 0$

$(x + 6)(x - 2) = 0$

$x + 6 = 0$

$x = 0 - 6$

$x {}^{1} = - 6$

$x - 2 = 0$

$x = 0 + 2$

$x {}^{2} = 2$

E)

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

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

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

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

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

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

$x(x - 1)4(x - 1) = 0$

$x - 1 = 0$

$x = 0 + 1$

$x {}^{1} = 1$

$x - 4 = 0$

$x = 0 + 4$

$x {}^{2} = 4$

F)

$(x - 3)(x + 2) = - 4$

$x {}^{2} + 2x - 3x - 6 = - 4$

$x {}^{2} - x - 6 = - 4$

$x {}^{2} - x - 6 + 4 = 0$

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

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

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

$x + 1 = 0$

$x = 0 - 1$

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

$x - 2 = 0$

$x = 0 + 2$

$x {}^{2} = 2$

G)

$x(x + 5) - 2x = 28$

$x {}^{2} + 5x - 2x = 28$

$x {}^{2} + 3x = 28$

$x {}^{2} + 3x - 28 = 0$

$x {}^{2} + 7x - 4x - 28 = 0$

$x(x + 7) - 4(x + 7) = 0$

$x + 7 = 0$

$x = 0 - 7$

$x {}^{1} = - 7$

$x - 4 = 0$

$x = 0 + 4$

$x {}^{2} = 4$

H)

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

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

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

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

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

$x(x + 5) - 4(x + 5)$

$x + 5 = 0$

$x = 0 - 5$

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

$x - 4 = 0$

$x = 0 + 4$

$x {}^{2} = 4$

I)

$2x(x + 3) = x {}^{2} + 3x + 70$

$2x {}^{2} + 6x = x {}^{2} + 3x + 70$

$2x {}^{2} + 6x - x {}^{2} - 3x - 70 = 0$

$x {}^{2} + 3x - 70 = 0$

$x {}^{2} + 10x - 7x - 70 = 0$

$x(x + 10) - 7(x + 10) = 0$

$x + 10 = 0$

$x = 0 - 10$

$x {}^{1} = - 10$

$x - 7 = 0$

$x = 0 + 7$

$x {}^{2} = 7$

J)

$(x + 3)(x - 4) - 52 = - x$

$x {}^{2} - 4x + 3x - 12 - 52 = - x$

$x {}^{2} - x - 64 = - x$

$x {}^{2} - x + x - 64 = 0$

$x {}^{2} - 64 = 0$

$x {}^{2} = 0 + 64$

$x {}^{2} = 64$

$x = \frac{ + }{} \sqrt{64}$

$x = \frac{ + }{} 8$

S=>{ +8 e -8}