Question 98541
{{{sqrt(7x+29) = x + 3 }}} Start with the given equation


{{{7x+29 = (x + 3)^2 }}} Square both sides



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



{{{0 = x^2-x-20 }}} Get all terms to one side



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{x^2-x-20=0}}} ( notice {{{a=1}}}, {{{b=-1}}}, and {{{c=-20}}})


{{{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  (note: remember when you square -1, you must square the negative as well. This is because {{{(-1)^2=-1*-1=1}}}.)




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




{{{x = (1 +- sqrt( 81 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (1 +- 9)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (1 +- 9)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (1 + 9)/2}}} or {{{x = (1 - 9)/2}}}


Lets look at the first part:


{{{x=(1 + 9)/2}}}


{{{x=10/2}}} Add the terms in the numerator

{{{x=5}}} Divide


So one answer is

{{{x=5}}}




Now lets look at the second part:


{{{x=(1 - 9)/2}}}


{{{x=-8/2}}} Subtract the terms in the numerator

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our possible solutions are:

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



Check:

Lets check the possible solution x=5

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


{{{sqrt(7(5)+29) = 5 + 3 }}} Plug in x=5


{{{sqrt(35+29) = 5 + 3 }}} Multiply


{{{sqrt(64) = 8 }}} Add


{{{8 = 8 }}} Take the square root of both sides



Since the equation is true, x=5 is a solution


------------------------------

Lets check the possible solution x=-4

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


{{{sqrt(7(-4)+29) = -4 + 3 }}} Plug in x=-4


{{{sqrt(-28+29) = -4 + 3 }}} Multiply


{{{sqrt(1) = -1 }}} Add


{{{1 = -1 }}} Take the square root of both sides


Since the equation is not true, x=-4 is not a solution



So our only solution is x=5