Question

Me ayudan por favor!!

Answer 1

Sea una ecuación cuadratica:

${ax}^{2} + bx + c = 0$

Se cumple:

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

a)

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

$x = \frac{ - ( - 1) \frac{ + }{} \sqrt{ {( - 1)}^{2} - 4(4)( - 3) } }{2(4)} \\ x = \frac{1 \frac{ + }{} \sqrt{1 + 48} }{8} \\ x = \frac{1 \frac{ + }{} \sqrt{49} }{8} \\ x = \frac{1 \frac{ + }{}7 }{8} \\ x = \frac{1 + 7}{8} \: \: \: o \: \: \: x = \frac{1 - 7}{8} \\ x = 1 \: \: \: o \: \: \: x = - \frac{3}{4}$

b)

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

$x = \frac{ - 5 \frac{ + }{} \sqrt{ {5}^{2} - 4(2)(2) } }{2(2)} \\ x = \frac{ - 5 \frac{ + }{} \sqrt{25 - 16} }{4} \\ x = \frac{ - 5 \frac{ + }{} \sqrt{9} }{4} \\ x = \frac{ - 5 \frac{ + }{}3 }{4} \\ x = \frac{ - 5 + 3}{4} \: \: \: o \: \: \: x = \frac{ - 5 - 3}{4} \\ x = - \frac{1}{2} \: \: \: o \: \: \: x = - 2$

c)

${3x}^{2} + 11x + 6 = 0$

$x = \frac{ - 11 \frac{ + }{} \sqrt{ {11}^{2} - 4(3)(6) } }{2(3)} \\ x = \frac{ - 11 \frac{ + }{} \sqrt{121 - 72} }{6} \\ x = \frac{ - 11 \frac{ + }{}7 }{6} \\ x = \frac{ - 11 + 7}{6} \: \: \: o \: \: \: x = \frac{ - 11 - 7}{6} \\ x = - 3 \: \: \: o \: \: \: x = - \frac{2}{3}$

d)

${6x}^{2} + 2 = 7x \\ {6x}^{2} - 7x + 2 = 0$

$x = \frac{ - ( - 7) \frac{ + }{} \sqrt{ {( - 7)}^{2} - 4(6)(2) } }{2(6)} \\ x = \frac{7 \frac{ + }{} \sqrt{49 - 48} }{12} \\ x = \frac{7 \frac{ + }{} 1}{12} \\ x = \frac{7 + 1}{12} \: \: \: o \: \: \: x = \frac{7 - 1}{12} \\ x = \frac{2}{3} \: \: \: o \: \: \: x = \frac{1}{2}$

e)

${5x}^{2} - 2x = 16 \\ {5x}^{2} - 2x - 16 = 0$

$x = \frac{ - ( - 2) \frac{ + }{} \sqrt{ {( - 2)}^{2} - 4(5)( - 16)} }{2(5)} \\ x = \frac{2 \frac{ + }{} 4 +320 }{10} \\ x = \frac{2 \frac{ + }{}18 }{10} \\ x = \frac{2 + 18}{10} \: \: \: o \: \: \: x = \frac{2 - 18}{10} \\ x = 2 \: \: \: o \: \: \: x = - \frac{8}{5}$

Answer 2

Respuesta:

Explicación paso a paso:

buen domingo