Question 1196337
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Square both sides to get rid of the square roots. They basically cancel out. 
It's like how multiplication and division are opposites one another.


{{{sqrt(x^2+5x) = sqrt(2x+4)}}}


{{{(sqrt(x^2+5x))^2 = (sqrt(2x+4))^2}}} Squaring both sides


{{{x^2+5x = 2x+4}}}


{{{x^2+5x-2x-4=0}}} Get everything to one side


{{{x^2+3x-4=0}}}


{{{(x+4)(x-1)=0}}} Factoring


{{{x+4=0}}} or {{{x-1=0}}} Zero product property


{{{x = -4 }}} or {{{x = 1}}} Solve each equation for x.


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The two *possible* solutions are 
{{{x = -4 }}} or {{{x = 1}}}


However, we need to check them both
Do so in the <font size=4>original equation</font> 


Let's check x = -4
{{{sqrt(x^2+5x) = sqrt(2x+4)}}}


{{{sqrt((-4)^2+5(-4)) = sqrt(2(-4)+4)}}}


{{{sqrt(16-20) = sqrt(-8+4)}}}


{{{sqrt(-4) = sqrt(-4)}}}
This x value leads to the same thing on both sides, but notice how the simplified radicand is negative.


The domain of {{{y = sqrt(x)}}} is {{{x >=0}}}, i.e. x is nonnegative, when we want y to be a real number. 
So it appears your teacher is ignoring complex/imaginary numbers here. 
You are absolutely correct that x = -4 is a solution but only in the complex number set. 
Regarding real numbers only, we have x = -4 to be extraneous.


Now check x = 1
{{{sqrt(x^2+5x) = sqrt(2x+4)}}}


{{{sqrt((1)^2+5(1)) = sqrt(2(1)+4)}}}


{{{sqrt(1+5) = sqrt(2+4)}}}


{{{sqrt(6) = sqrt(6)}}}
This time the radicands are positive, so we don't run into the same issue as earlier.
Both sides are the same real number (approximately 2.449), which confirms x = 1 is the solution.


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If you were to graph {{{y = sqrt(x^2+5x)}}} and {{{y = sqrt(2x+4)}}}, as shown in the Desmos link below,
<a href = "https://www.desmos.com/calculator/9n9fqdwu9p">https://www.desmos.com/calculator/9n9fqdwu9p</a>
Notice how the two curves cross at (1, sqrt(6)) and this is the only intersection point. This visually confirms x = 1 as the solution.
sqrt(6) = 2.449 approximately
There is no intersection point for when x = -4, which helps rule out this extraneous value.
In fact, -4 isn't in the domain of either graph.


Side note: Feel free to use other graphing apps like GeoGebra or a TI83/TI84 if you prefer that better.
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