Question

Question 11 of 25

Answer 1

Answer:

${ \rm{x + 4 = \sqrt{8 - x} }} \\ \\ { \rm{ {(x + 4)}^{2} = {( \sqrt{8 - x}) }^{2} }} \\ \\ { \rm{(x + 4)(x + 4) = 8 - x}} \\ \\ { \rm{ {x}^{2} + 8x + 16 = 8 - x}} \\ \\ { \rm{ {x}^{2} + 9x + 8 = 0 }} \\ \\ { \rm{(x + 1)(x + 8) = 0}} \\ \\ { \boxed{ \tt{x _{1} = {}^{ - } 1}}} \: \: { \rm{and}} \: \: { \boxed{ \tt{x _{2} = {}^{ - } 8}}}$

Answer 2

$\\ \sf\longmapsto x+4=\sqrt{8-x}$

$\\ \sf\longmapsto (x+4)^2=8-x$

$\\ \sf\longmapsto x^2+16+8x=8-x$

$\\ \sf\longmapsto x^2+8x-x+16-8=0$

$\\ \sf\longmapsto x^2+7x+8=0$

$\\ \sf\longmapsto x^2-8x+x+8$

$\\ \sf\longmapsto (x-8)(x-1)=0$

$\\ \sf\longmapsto x=8,-1$

Option C