Respuesta :

gvcci
do recall that squaring and the *radical sign* cancel each other out... like so:([tex] \sqrt{a} [/tex])[tex] ^{2} [/tex]= a

When you put it that way, it isn't enough :P
([tex] \sqrt{a} [/tex])[tex] ^{2} [/tex]= a
([tex] \sqrt{8x+1} [/tex])[tex] ^{2} [/tex]=?

so you start with
([tex] \sqrt{8x+1} [/tex])[tex] ^{2} [/tex]= [tex] (5)^{2} [/tex]
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3

To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?

(extraneous solution is a invalid solution)

x=3 not extraneous

Answer with explanation:

The given equation is

    [tex]\sqrt{8 x +1}=5\\\\ \text{Squaring both sides}}\\\\ 8x +1=25\\\\ 8 x=25-1\\\\8 x=24\\\\x=\frac{24}{8}\\\\x=3[/tex]

Substituting the value of , x=3, in original equation

LHS

   [tex]\rightarrow\sqrt{8\times 3 +1}\\\\\sqrt{25}=5[/tex]

=R HS,

So, x=3, is not an extraneous solution.

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