Respuesta:
a) √(x + 5) = -2x + 1
x + 5 = (-2x + 1)²
x + 5 = 4x² + 1 - 2•2x•1
x + 5 = 4x² + 1 - 4x
4x² + 1 - 4x - x - 5 = 0
4x² - 5x - 4 = 0
x = 5 ± √(25 + 4•4•4) / 2•4
x = 5 ± √(25 + 64) / 8
x = 5 ± √89 / 8
x1 = (5 + √89)/8
x2 = (5 - √89)/8
b) ½x + 2 = √(3x + 2)
(½x + 2)² = 3x + 2
x²/4 + 4 + 2x = 3x + 2
¼x² + 2x - 3x + 4 - 2 = 0
¼x² - x + 2 = 0
x² - 4x + 8 = 0
x = 4 ± √(16 - 4•8) / 2
x = 4 ± √(16 - 32) / 2
x = 4 ± √-16 / 2
x1 = 4 + √-16 /2
x2 = 4 - √-16 /2
No tiene soluciones en los reales
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Respuesta:
a) √(x + 5) = -2x + 1
x + 5 = (-2x + 1)²
x + 5 = 4x² + 1 - 2•2x•1
x + 5 = 4x² + 1 - 4x
4x² + 1 - 4x - x - 5 = 0
4x² - 5x - 4 = 0
x = 5 ± √(25 + 4•4•4) / 2•4
x = 5 ± √(25 + 64) / 8
x = 5 ± √89 / 8
x1 = (5 + √89)/8
x2 = (5 - √89)/8
b) ½x + 2 = √(3x + 2)
(½x + 2)² = 3x + 2
x²/4 + 4 + 2x = 3x + 2
¼x² + 2x - 3x + 4 - 2 = 0
¼x² - x + 2 = 0
x² - 4x + 8 = 0
x = 4 ± √(16 - 4•8) / 2
x = 4 ± √(16 - 32) / 2
x = 4 ± √-16 / 2
x1 = 4 + √-16 /2
x2 = 4 - √-16 /2
No tiene soluciones en los reales