Question 758352

{{{sqrt (x+2) + sqrt (3x+7) = 1}}}....square both sides


{{{(sqrt (x+2))^2 + 2sqrt (x+2)sqrt (3x+7)+(sqrt (3x+7))^2 = 1^2}}}


{{{x+2 + 2sqrt((x+2)(3x+7))+3x+7 = 1}}}

{{{2sqrt((x+2)(3x+7))+4x+9 = 1}}}

{{{2sqrt((x+2)(3x+7))=-4x-9+ 1}}}

{{{2sqrt((x+2)(3x+7))=-4x-8}}}

{{{2sqrt((x+2)(3x+7))/2=-4x/2-8/2}}}

{{{sqrt((x+2)(3x+7))=-2x-4}}}...square both sides

{{{(sqrt((x+2)(3x+7)))^2=(-2x-4)^2}}}

{{{(x+2)(3x+7)=4x^2+16x+16}}}

{{{3x^2+7x+6x+14=4x^2+16x+16}}}

{{{3x^2+13x+14=4x^2+16x+16}}}

{{{0=4x^2-3x^2+16x-13x+16-14}}}

{{{x^2+3x+2=0}}}...write {{{3x}}} as {{{2x+x}}}

{{{x^2+2x+x+2=0}}}.....group

{{{(x^2+x)+(2x+2)=0}}}

{{{x(x+1)+2(x+1)=0}}}

{{{(x+1)(x+2)=0}}}.......use zero product rule

if {{{x+1=0}}} => {{{x=-1}}}

if {{{x+2=0}}} => {{{x=-2}}}

 only solution that works is {{{highlight(x=-2)}}} because it makes term {{{sqrt (x+2)}}} equal to zero and term {{{sqrt (3x+7)}}} equal to {{{1}}}