Question 195655
{{{sqrt(7x+29)= x+3}}} Start with the given equation.



{{{7x+29=(x+3)^2}}} Square both sides (to eliminate the square root)



{{{7x+29=x^2+6x+9}}} FOIL



{{{0=x^2+6x+9-7x-29}}} Get everything to one side



{{{0=x^2-x-20}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-1}}}, and {{{c=-20}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(-20) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-1}}}, and {{{c=-20}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(-20) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(-20) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--80 ))/(2(1))}}} Multiply {{{4(1)(-20)}}} to get {{{-80}}}



{{{x = (1 +- sqrt( 1+80 ))/(2(1))}}} Rewrite {{{sqrt(1--80)}}} as {{{sqrt(1+80)}}}



{{{x = (1 +- sqrt( 81 ))/(2(1))}}} Add {{{1}}} to {{{80}}} to get {{{81}}}



{{{x = (1 +- sqrt( 81 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- 9)/(2)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{x = (1 + 9)/(2)}}} or {{{x = (1 - 9)/(2)}}} Break up the expression. 



{{{x = (10)/(2)}}} or {{{x =  (-8)/(2)}}} Combine like terms. 



{{{x = 5}}} or {{{x = -4}}} Simplify. 



So the <i>possible</i> solutions are {{{x = 5}}} or {{{x = -4}}} 



However, if you plug in {{{x = -4}}} back into the original equation, you'll get a contradiction.


So {{{x = -4}}} is NOT a solution



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Answer: 



So the solution is {{{x = 5}}}