Answer:
x=8 is a extraneous solution.
Step-by-step explanation:
We have: [tex]\sqrt{4x+41} =x + 5[/tex]
→ [tex]4x + 41 =(x+5)^{2}[/tex]
→ [tex]4x + 41 =(x)^{2}+ 10x + 25[/tex]
→ [tex]0 =(x)^{2} + 6x -16[/tex]
Factorizing we have that:
→ [tex](x+8)(x-2)[/tex]
Therefore, we have two solutions:
x1=-8 and x2=2
Then, we have that when x=-8:
[tex]\sqrt{4(-8)+41} = -8 + 5[/tex]
[tex]\sqrt{4(-8)+41} = -3[/tex]
Therefore, like the result of the square root of the number should be negative, x=8 is a extraneous solution.
