Question

-4x(2+x)=8 solve equation
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Answer 1

$- 4x(2 + x) = 8 \\ - x(2 + x) = 2 \\ - 2x - {x}^{2} = 2 \\ {x}^{2} + 2x + 2 = 0 \\ {x}^{2} + 2x + 1 - 1 = - 2 \\ {x}^{2} + 2x + 1= - 1 \\ {(x + 1)}^{2} = - 1 \\ x + 1 = \sqrt{ - 1} \\ x = - 1 + \sqrt{ - 1} \\ or \: x = - 1 - \sqrt{ - 1}$

i.e., x = -1-i or x = -1+i

i.e., there's no real values for x.

Answer 2

-4x(2 + x) = 8

-8x - 4x² - 8 = 0

4x² + 8x + 8 = 0

b² - 4ac = 64 - 128 = -64 < 0

since the discriminant is smaller than zero, the roots are imaginary.

(if you are in 11th and have learnt the chapter "complex numbers", only then the further part is applicable for you to apply)

$x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \: \: \: \: \: \: or \: \: \: \: \: \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - 8 + \sqrt{ - 64} }{8} \: \: \: \: or \: \: \: \: \frac{ - 8 - \sqrt{ - 64} }{8} \\ x = \frac{ - 8 + \sqrt{(64)( - 1)} }{8} \: or \: \frac{ - 8 - \sqrt{(64)( - 1)} }{8} \\ x = \frac{ - 8 + 8 \sqrt{ - 1} }{8} \: or \: \frac{ - 8 - 8 \sqrt{ - 1} }{8} \\ x = \frac{ - 8 + 8i}{8} \: \: \: \: or \: \: \: \: \frac{ - 8 - 8i}{8} \\ x = - 1 + i \: \: \: \: or \: \: \: \: - 1 - i$

Hence, the solutions of the given equation are either x = -1 + i

or x = -1 - i.