Answer:
No
Edit:
Yes, based on original equation. (Credit to greenpumpkin for correction)
Explanation:
For this problem, we simply need to find the values of x that can make the equation true. So, let's begin by isolating the "x" variable.
sqrt(2x + 13) = x + 5
[sqrt(2x + 13)]^2 = (x + 5)^2
2x + 13 = x^2 + 10x + 25
0 = x^2 + 8x + 12
Note, we can remove the sqrt method by squaring both sides of the equation. Doing this, we see we have a quadratic equation meaning we can apply the quadratic formula to find solutions for x.
[-b +/- sqrt( b^2 - 4(a)(c) ) ] / 2a
Let a = 1, b = 8, and c = 12
[-8 +/- sqrt( (8)^2 - 4(1)(12) ) ] / 2(1)
= [-8 +/- sqrt( 64 - 48 ) ] / 2
= [-8 +/- sqrt(16) ] / 2
= [ -8 +/- 4 ] / 2
So, x = [ -8 + 4 ] / 2 and x = [-8 - 4 ] / 2
x = [-4] / 2 = -2 and x = [-12] / 2 = -6
Hence, the two values of x that can solve this quadratic equation are x = -2 and x = -6.
Therefore, we know that x = -6 is not extraneous, meaning it is a solution to our equation.
Cheers.
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Edit:
Plugging the value of -6 back into the original equation, we get the following:
sqrt(2x + 13) = x + 5
sqrt(2(-6) + 13) = (-6) + 5
sqrt (1) = -1
1 != -1
Given that 1 cannot equal negative 1, we can say that x = -6 is an extraneous solution. (Credit to greenpumpkin for correction)