Answer:
x=-8
Step-by-step explanation:
[tex]\sqrt{4x+41}=x+5[/tex]
on squaring both sides
[tex]4x+41=(x+5)^2\\\\4x+41=x^2+10x+25\\\\x^2+6x-16=0\\\\x^2+8x-2x-16=0\\\\x(x+8)-2(x+8)=0\\\\(x-2)(x+8)=0[/tex]
either x-2=0 or x+8=0
either x=2 or x=-8
Putting x=2 in original equation, we get
[tex]\sqrt{4\times 2+41}=2+5[/tex]
7=7 hence, it is not an extraneous solution
Putting x=-8 in original equation
[tex]\sqrt{4\times (-8)+41}=-8+5[/tex]
i.e. 3=-3
Hence, x=-8 is an extraneous solution (since it doesn't satisfy the original equation)
In this equation, the answer to 4x + 41 = x + 5 is:
It is a factor that when multiplied by itself gives the original number.
√4x + 41 = x + 5.
On squaring both sides will give:
4x +41 = (x + 5)²
4x + 41= x² + 10x + 25
x² + 6x -16= 0
x² - 8x - 2x - 16 = 0
x (x + 8) - 2(x + 8) = 0
(x-2) (x+ 8)= 0
Therefore: either x-2=0 or x+8=0
either x=2 or x=-8
Putting x=2 in original equation, we get:
√4 X 2+ 41 =2 + 5
7=7 hence, it is not an extraneous solution
Putting x=-8 in original equation
√4 X (-8) + 41 = -8 + 5
That is 3=-3
Hence, x=-8 is an extraneous solution (since it doesn't satisfy the original equation)
x = -8
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