SOLUTION: [ I am reposting because there was a typo in my email when I posted the previous question ] Could you please help me with this equation? The solution manual says the answer is x=1

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Question 1196337: [ I am reposting because there was a typo in my email when I posted the previous question ]
Could you please help me with this equation? The solution manual says the answer is x=1 but I am getting x = 1 and x= - 4.
Could you please help me understand why - 4 is not a solution? I checked it and it doesn't seem to be an extraneous solution. Would be a huge help if you could please include an explanation for the steps.
Would greatly appreciate your help.
IMG-4646
Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
DUPLICATE

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NOT one bit different than your other post. Try substituting -4 for x, and find out!!!

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

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%28x%5E2%2B5x%29+=+sqrt%282x%2B4%29

%28sqrt%28x%5E2%2B5x%29%29%5E2+=+%28sqrt%282x%2B4%29%29%5E2 Squaring both sides

x%5E2%2B5x+=+2x%2B4

x%5E2%2B5x-2x-4=0 Get everything to one side

x%5E2%2B3x-4=0

%28x%2B4%29%28x-1%29=0 Factoring

x%2B4=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 original equation

Let's check x = -4
sqrt%28x%5E2%2B5x%29+=+sqrt%282x%2B4%29

sqrt%28%28-4%29%5E2%2B5%28-4%29%29+=+sqrt%282%28-4%29%2B4%29

sqrt%2816-20%29+=+sqrt%28-8%2B4%29

sqrt%28-4%29+=+sqrt%28-4%29
This x value leads to the same thing on both sides, but notice how the simplified radicand is negative.

The domain of y+=+sqrt%28x%29 is x+%3E=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%28x%5E2%2B5x%29+=+sqrt%282x%2B4%29

sqrt%28%281%29%5E2%2B5%281%29%29+=+sqrt%282%281%29%2B4%29

sqrt%281%2B5%29+=+sqrt%282%2B4%29

sqrt%286%29+=+sqrt%286%29
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%28x%5E2%2B5x%29 and y+=+sqrt%282x%2B4%29, as shown in the Desmos link below,
https://www.desmos.com/calculator/9n9fqdwu9p
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.