Question

Banda quien me ayuda con mi tarea? Hagan paro solo gente honesta xfa :(

Answer 1

Respuesta:

Hola... Espero sea de ayuda...

Problema #1⇩

${x}^{2} - 14x + 45 = 0 \\ \\ x = - b \frac{ + }{ - } \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = -( - 14) \frac{ + }{ - } \frac{ \sqrt{ { - 14}^{2} - 4(1)(45)} }{2(1)} \\ \\ x = 14 \frac{ + }{ - } \frac{ \sqrt{ 196 -180} }{2} \\ \\ x = 14 \frac{ + }{ - } \frac{ \sqrt{16} }{2} \\ \\ x = 14 \frac{ + }{ - } \frac{ 4 }{2} \\ \\ x = 14 + \frac{4}{2} \\ x = \frac{18}{2} = 9 \\ \\ x = 14 - \frac{4}{2} = \frac{10}{2} = 5$

Problema #2⇩

$2 {x}^{2} - x - 1 = 0 \\ \\ x = - b \frac{ + }{ - } \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = - 1 \frac{ + }{ - } \frac{ \sqrt{( - 1 )^{2} - 4(2)( - 1) } }{2( - 1)} \\ \\ x = - 1 \frac{ + }{ - } \frac{ \sqrt{1 + 8} }{ - 2} \\ \\ x = - 1 \frac{ + }{ - } \frac{ \sqrt{9} }{ - 2} \\ \\ x = - 1 \frac{ + }{ - } \frac{3}{ - 2} \\ \\ x = - 1 + \frac{3}{ - 2} \\ x = \frac{2}{ - 2} \\ x = - 1 \\ \\ x = - 1 - \frac{3}{ - 2} \\ x = \frac{ - 4}{ - 2} \\ x = 2$

Problema #3⇩

$6 {x}^{2} + 5x + 1 = 0\\ \\ x = - b \frac{ + }{ - } \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = - 5 \frac{ + }{ - } \frac{ \sqrt{ {5}^{2} - 4(6)(1)} }{2(6)} \\ \\ x = - 5 \frac{ + }{ - } \frac{ \sqrt{ {25} - 24} }{12} \\ \\ x = - 5 \frac{ + }{ - } \frac{ \sqrt{ 1} }{12} \\ \\ x = - 5 \frac{ + }{ - } \frac{1}{12} \\ \\ x = - 5 + \frac{1}{ 12} \\ x = \frac{ - 4}{12} \\ x = - \frac{1}{3} \\ \\ x = - 5 - \frac{1}{ 12} \\ x = \frac{ - 6}{12} \\ x = - \frac{1}{2}$

Problema #4⇩

$2 {x}^{2} - 3x - 9 = 0 \\ \\ x = - b \frac{ + }{ - } \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = - ( - 3) \frac{ + }{ - } \frac{ \sqrt{ { - 3}^{2} - 4(2)( - 9)} }{2(2)} \\ \\ x = 3 \frac{ + }{ - } \frac{ \sqrt{ 9 + 72} }{2(2)} \\ \\ x = 3\frac{ + }{ - } \frac{ \sqrt{81} }{4} \\ \\ x = 3 \frac{ + }{ - } \frac{9}{4} \\ \\ x = 3 + \frac{9}{4 } \\ x = \frac{12}{4} \\ x = 3 \\ \\ x = 3 - \frac{9}{4} = \frac{ - 6}{4} = - \frac{3}{2}$

Problema #5⇩

$4 {x}^{2} - 9x + 3 = 0 \\ \\ x = - b \frac{ + }{ - } \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = - ( - 9) \frac{ + }{ - } \frac{ \sqrt{ { - 9}^{2} - 4(4)(3)} }{2(4)} \\ \\ x = 9 \frac{ + }{ - } \frac{ \sqrt{ 81 - 48} }{8} \\ \\ x = 9\frac{ + }{ - } \frac{ \sqrt{ 33} }{8} \\ \\ x = 9 + \frac{ \sqrt{33} }{8} \\ \\ x = 9 - \frac{ \sqrt{33} }{8}$

Answer 2

Explicación paso a paso:

El de arriba está bien.