Question

halla la solucion de cada ecuacion porfabor se los pido de corazon

se los pido pero con solucion​.

Answer 1

${x}^{2} + 16 = 4 \\ {x}^{2} + 16 - 4 \\ {x}^{2} + 12 = 0 \\ {x}^{2} = - 12 \\ x = \sqrt{ - 12}$

sin solución

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${5x}^{2} + 40 = - 60 \\ {5x}^{2} = - 60 - 40 \\ {5x}^{2} = - 100 \\ {x}^{2} = - 100 \div 5 \\ {x}^{2} = - 20 \\ x = \sqrt{ - 20}$

sin solución

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${ \frac{1}{6}r }^{2} - { \frac{1}{6}r }^{2} = \frac{2}{3} \\ \\ \frac{{ - \frac{1}{6}r }^{2} }{ - \frac{ 1}{6} } = \frac{ \frac{2}{3} }{ - \frac{1}{6} } \\ \\ {r}^{2} = 4 \\ \\ r = \sqrt{4} \\ \\ r = 2$

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$- {y}^{2} + 8 = - {5y}^{2} \\ - {y}^{2} + 8+ ( - {5y}^{2} ) = 0 \\ - {4y}^{2} + 8 = 0 \\ - {4y}^{2} = - 8 \\ {y}^{2} = - 8 \div 4 \\ {y}^{2} = - 2 \\ y = \sqrt{ - 2} </p><p>$

sin solución

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$57 = 30 - {x}^{2} \\ 57 - 30 = - {x}^{2} \\ 27 = - {x}^{2} \\ - {x}^{2} = 27 \\ {x}^{2} = - 27 \\ x = \sqrt{ - 27}$

sin solución

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$- \frac{3}{15} {x}^{2} = - \frac{2}{5} {x}^{2} - \frac{12}{5} \\ \\ - \frac{1}{5} {x}^{2} = - \frac{2}{5} {x}^{2} - \frac{12}{5} \\ \\ - \frac{1}{5} {x}^{2} + \frac{2}{5} {x}^{2} = - \frac{12}{5} \\ \\ \frac{1}{5} {x}^{2} = - \frac{12}{5} \\ \\ {x}^{2} = - \frac{12}{5} \div \frac{1}{5} \\ \\ {x}^{2} = - 12 \\ \\ x = \sqrt{ - 12}$

sin solución

Nota : La raíz cuadrada de un número negativo no es un número real