Question

ayudaaaaaaa porfavoooor​

Answer 1

Esta es la formula general:

Sea:

$a {x}^{2} + bx + c = 0$

Se cumple:

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

Ahora solo hay que reemplazar los coeficientes en la formula:

a)

$3 {x}^{2} - 5x + 2 = 0$

...

$x = \frac{ - ( - 5) \frac{ + }{} \sqrt{ {( - 5)}^{2} - 4(3)(2)} }{2(3)} \\ x = \frac{5 \frac{ + }{} \sqrt{25 - 24} }{6} \\ x = \frac{5 \frac{ + }{} \sqrt{1} }{6} \\ x = \frac{5 \frac{ + }{}1 }{6} \\ x = \frac{5 + 1}{6} \: \: \: o \: \: \: x = \frac{5 - 1}{6} \\ x = 1 \: \: \: o \: \: \: x = \frac{2}{3}$

b)

${x}^{2} + 11x + 24 = 0$

...

$x = \frac{ - 11 \frac{ + }{} \sqrt{ {11}^{2} - 4(1)(24)} }{2(1)} \\ x = \frac{ - 11 \frac{ + }{} \sqrt{121 - 96} }{2} \\ x = \frac{ - 11 \frac{ + }{} \sqrt{25} }{2} \\ x = \frac{ - 11 \frac{ + }{}5 }{2} \\ x = \frac{ - 11 + 5}{2} \: \: \: o \: \: \: x = \frac{ - 11 - 5}{2} \\ x = - 3 \: \: \: o \: \: \: x = - 8$

c)

$9 {x}^{2} - 12x + 4 = 0$

...

$x = \frac{ - ( - 12) \frac{ + }{} \sqrt{ {( - 12)}^{2} - 4(9)(4)} }{2(9)} \\ x = \frac{12 \frac{ + }{} \sqrt{144 - 144} }{18} \\ x = \frac{12 \frac{ + }{} \sqrt{0} }{18} \\ x = \frac{12 \frac{ + }{} 0}{18} \\ x = \frac{12}{18} \\ x = \frac{2}{3}$

d)

$6 {x}^{2} = x + 222 \\ 6 {x}^{2} - x - 222 = 0$

...

$x = \frac{ -( - 1) \frac{ + }{} \sqrt{ {(1)}^{2} - 4(6)( - 222)} }{2(6)} \\ x = \frac{1 \frac{ + }{} \sqrt{1 + 5328} }{12} \\ x = \frac{1 \frac{ + }{} \sqrt{5329} }{12} \\ x = \frac{1 + 73}{12} \\ x = \frac{74}{12} \: \: \: o \: \: \: x = \frac{ - 72}{12} \\ x = \frac{37}{6} \: \: \: o \: \: \: x = - 6$

e)

$8x + 5 = 36 {x}^{2} \\ 36 {x}^{2} - 8x - 5 = 0$

...

$x = \frac{ - 8 \frac{ + }{} \sqrt{ {8}^{2} - 4(36)( - 5)} }{2(36)} \\ x = \frac{ - 8 \frac{ + }{} \sqrt{64 + 720} }{72} \\ x = \frac{ - 8 \frac{ + }{} \sqrt{784} }{72} \\ x = \frac{ - 8 \frac{ + }{} 28}{72} \\ x = \frac{ - 8 + 28}{72} \: \: \: o \: \: \: x = \frac{ - 8 - 28}{72} \\ x = \frac{5}{18} \: \: \: o \: \: \: x = - \frac{1}{2}$