Stephanie is student teaching at her local high school. She has to help her instructor write a test. She has to write a radical equation where the solution is extraneous. She also has to write a radical equation where the solution is non-extraneous. Help Stephanie write one radical equation where the solution is extraneous and another equation where the solution is non-extraneous. Using complete sentences, explain each step when solving to justify your examples.
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OpenStudy (anonymous):
I made the non extraneous solution but the extraneous solution i'm getting confused about.
OpenStudy (anonymous):
x^2+4=81
OpenStudy (anonymous):
why negative?
OpenStudy (precal):
sorry my computer is not cooperating
OpenStudy (anonymous):
it's fine and x^2 and 4 are square roots
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OpenStudy (precal):
\[\sqrt{7x-1}+8=0\]
OpenStudy (precal):
\[\sqrt{7x-1}+8=0\]
OpenStudy (precal):
ok this one will work, let me explain
OpenStudy (precal):
subtract 8 to both sides
OpenStudy (anonymous):
square both sides
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OpenStudy (precal):
\[\sqrt{7x-1}=-8\]
OpenStudy (anonymous):
7x-1=64
OpenStudy (precal):
finish solving it
OpenStudy (anonymous):
x=9
OpenStudy (precal):
no
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OpenStudy (anonymous):
7x=65
OpenStudy (precal):
7x-1=64
+1 +1
7x=65
divide 7 to both sides
x=65/7
OpenStudy (precal):
ok now sub into original equation
OpenStudy (anonymous):
sry i subtracted instead of added
OpenStudy (precal):
no problem, test it
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OpenStudy (anonymous):
\[\pm 8 = 8\]
OpenStudy (anonymous):
-8 i meant
OpenStudy (precal):
no
OpenStudy (precal):
\[\sqrt{7\left( \frac{ 65 }{ 7 } \right)-1}+8=0\]
OpenStudy (precal):
\[\sqrt{65-1}+8=0\]
\[\sqrt{64}+8=0\]
8+8=0
16=0
false therefore solution is extraneous
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OpenStudy (anonymous):
oh, thx, may i have the original solution
OpenStudy (precal):
x=65/7
OpenStudy (precal):
ok use the one I left
OpenStudy (anonymous):
And here's a medal
OpenStudy (precal):
thanks hope that help you
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OpenStudy (anonymous):
You're welcome
OpenStudy (anonymous):
May i ask is This an extraneous solution \[\sqrt{x}=3\]
OpenStudy (precal):
no because the square root of 9 is 3 so x=9 is a solution
OpenStudy (anonymous):
so it's non-extraneous
OpenStudy (precal):
its a solution it is not extraneous
its either extraneous or a solution
extraneous-extra solution that is in error (it doesn't work)
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