Question

$\sqrt[2]{x+2} + \sqrt[2]{x+8} =1$

Answer 1

Respuesta:

$\sqrt{x + 2} + \sqrt{x + 8} = 1$

Elevas al cuadrado a cada lado.

$( \sqrt{x + 2} + \sqrt{x + 8})^{2} = {1}^{2}$

$( \sqrt{x + 2})^{2} + 2 \times( \sqrt{x + 2}) \times (\sqrt{x + 8}) + (\sqrt{x + 8})^{2} = 1$

$x + 2 + 2 \sqrt{(x + 2)(x + 8)} + x + 8 = 1$

$2x + 10 + 2 \sqrt{(x + 2)(x + 8)} = 1$

$2 \sqrt{(x + 2)(x + 8)} = - 2x - 9$

$\sqrt{(x + 2)(x + 8)} = ( - 2x - 9) \div 2$

$\sqrt{ {x}^{2} + 10x + 16} = ( - 2x - 9) \div 2$

De nuevo elevas al cuadrado a cada lado

$( \sqrt{ {x}^{2} + 10x + 16})^{2} = (( - 2x - 9) \div 2)^{2}$

${x}^{2} + 10x + 16 = ( 4 {x}^{2} + 2 \times 2x \times 9 + 81 ) \div 4$

${x}^{2} + 10x + 16 = {x}^{2} + 9x + \frac{81}{4}$

${x}^{2} + 10x + 16 - {x}^{2} - 9x - \frac{81}{4} = 0$

$x + 16 - \frac{81}{4} = 0$

$x = \frac{81}{4} - 16$

$x = (81 - 64) \div 4$

$x = \frac{17}{4}$