Question 84334
Square root of x+56=x

{{{sqrt(x+56)=x}}} square both sides to get rid of the radical and we get:

{{{x+56=x^2}}}  subtract x and also 56 from both sides

{{{x^2-x-56=x-x+56-56}}}  collect like terms

{{{x^2-x-56=0}}}  quadratic in standard form and it can be factored

{{{(x-8)(x+7)=0}}}

{{{x-8=0}}}

{{{x=8}}} and

{{{x+7=0}}}

{{{x=-7}}}

CK

{{{sqrt(8+56)=8}}}
{{{sqrt(64)=8}}}
{{{8=8}}}

{{{sqrt(-7+56)=-7}}}
{{{sqrt (49)=-7}}}
Since the square root of 49 is plus or minus 7, this solution works also.

(Note: If the A coefficient of a quadratic is 1, then the B coefficient is the sum of the factors of the C coefficient.  In this case, the factors of the C coefficient that work are -8 and +7)


Hope this helps---ptaylor