Question 877641
Raise both sides to power of 2.
Simplify.


You found x=4 or x=-7 but the key said just x=4.

Let us see what we can find.


{{{(sqrt(29-x))^2=(x+1)^2}}}
{{{29-x=x^2+2x+1}}}
{{{0=x^2+2x+1-29+x}}}
{{{x^2+3x-28=0}}}---- Is that left member factorable?
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(x+4)(x-7), this one does not work;
(x-4)(x+7), this one works;
{{{(x-4)(x+7)=0}}}
One of the factors is zero or the other factor is zero.
x=4, OR x=-7
Both solutions will work.  


CHECK:  
{{{sqrt(29-4)=4+1}}}
{{{sqrt(25)=5}}}, Yes, x=4 works.
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{{{sqrt(29-(-7))=-7+1}}}
{{{sqrt(36)=-6}}}
{{{sqrt(6*6)=6}}} or-inclusive {{{sqrt(6*6)=-6}}} actually; so yes, this x=-7 also will work.  
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Both solutions are correct.  
{{{6*6=36}}} AND {{{(-6)(-6)=36}}}