Question 627353
{{{sqrt(2x+11) = x+7}}}


{{{2x+11 = (x+7)^2}}}


{{{2x+11 = x^2+14x+49}}}


{{{0 = x^2+14x+49-2x-11}}}


{{{0 = x^2+12x+38}}}


{{{x^2+12x+38 = 0}}}


Now let's use the quadratic formula for x


{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}}


{{{x = (-(12) +- sqrt( (12)^2-4(1)(38) ))/(2(1))}}} 


{{{x = (-12 +- sqrt( -8 ))/(2)}}}


Since you can't take the square root of a negative number, this means that there are no real solutions to {{{sqrt(2x+11) = x+7}}}.

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{{{sqrt(2x)+11 = x+7}}}


{{{sqrt(2x) = x+7-11}}}


{{{sqrt(2x) = x-4}}}


{{{2x = (x-4)^2}}}


{{{2x = x^2 - 8x + 16}}}


{{{0 = x^2 - 8x + 16-2x}}}


{{{0 = x^2 - 10x + 16}}}


{{{x^2 - 10x + 16 = 0}}}


Now let's use the quadratic formula for x


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


{{{x = (-(-10)+-sqrt((-10)^2-4(1)(16)))/(2(1))}}}


{{{x = (10+-sqrt(100-(64)))/(2)}}}


{{{x = (10+-sqrt(36))/2}}}


{{{x = (10+sqrt(36))/2}}} or {{{x = (10-sqrt(36))/2}}}


{{{x = (10+6)/2}}} or {{{x = (10-6)/2}}}


{{{x = 16/2}}} or {{{x = 4/2}}}


{{{x = 8}}} or {{{x = 2}}}


So the possible solutions to {{{sqrt(2x)+11 = x+7}}} are {{{x = 8}}} or {{{x = 2}}}


Make sure you check them (I'll let you do this)