Question

me podrian ayudar com esto ecuaciones radicales​

Answer 1

Explicación paso a paso:

a)

${( \sqrt{x + 3} = 4) }^{2} = x + 3 = 16 \\ x = 16 - 3 \\ x = 13$

C)

$5 - \sqrt{3x + 1} = 0 \\ { ( - \sqrt{3x + 1} = - 5) }^{2} \\ 3x + 1 = 25 \\ 3x = 25 - 1 \\ 3x = 24 \\ x = \frac{24}{3} \\ x = 8$

E)

$7 - \sqrt{x + 2} = 3 \\ - \sqrt{x + 2 } = 3 - 7 \\ {( - \sqrt{x + 2} = - 4) }^{2} \\ x + 2 = 16 \\ x = 16 - 2 \\ x = 14$