Question

resuelve las siguientes ecuaciones utilizando la ecuacion cuadratica
a(x+1)(x-5)=16
b(x+1)(x+4)=4
c)(x-2)(x-3)=4

Answer 1

primero tener la ecuacion cuadratica a mano

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$

a)

$(x + 1)(x - 5) = {x}^{2} - 5x + x - 5 = {x }^{2} - 4x - 5 = 16 \\ \\ {x}^{2} - 4x - 5 - 16 = {x}^{2} - 4x - 21 = 0 \\ \\ a = 1 \\ b = - 4 \\ c = - 21 \\ \\ x = \frac{4 + - \sqrt{ { - 4}^{2} - (4 \times 1 \times - 21) } }{2} = \frac{4 + - \sqrt{16 + 84} }{2 }= \frac{4 + - \sqrt{100} }{2 } \\ \\ x1 = \frac{4 + 10}{2} = \frac{14}{2} = 7 \\ \\ x2 = \frac{4 - 10}{2} = \frac{ - 6}{2} = - 3$

b)

$(x + 1)(x + 4) = {x}^{2} + 4x + x + 4 = {x}^{2} + 5x + 4 = 4 \\ \\ {x}^{2} + 5x + 4 - 4 = 0 \\ a = 1 \\ b = 5 \\ c = 0 \\ \\ x = \frac{ - 5 + - \sqrt{ {5}^{2} - (4 \times 1 \times 0) } }{2} = \frac{ - 5 + - \sqrt{25} }{2} \\ \\ x1 = \frac{ - 5 + \sqrt{25} }{2} = \frac{ - 5 + 5}{2} = 0 \\ x2 = \frac{ - 5 - \sqrt{25} }{2} = \frac{ - 5 - 5}{2} = \frac{ - 10}{2} = - 5$

c)

$(x - 2)(x - 3) = {x}^{2} - 3x - 2x + 6 = {x}^{2} - 5x + 6 = 4 \\ {x}^{2} - 5x + 6 - 4 = 0 \\ \\ a = 1 \\ b = - 5 \\ c = 2 \\ \\ x = \frac{ 5 + - \sqrt{ { - 5}^{2} - (4 \times 1 \times 2)} }{2} = \frac{5 + - \sqrt{25 - 8} }{2} = \frac{5 + - \sqrt{17} }{2 } \\ \\ x1 = \frac{5 + \sqrt{17} }{2} \\ x2 = \frac{5 - \sqrt{17} }{2}$