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OpenStudy (anonymous):
Find the solution of five times the square root of the quantity of x plus 7 equals negative 10 and determine if it is an extraneous solution. x = −3; extraneous x = −3; not extraneous x = 11; extraneous x = 11; not extraneous
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OpenStudy (anonymous):
-3 and extraneous
OpenStudy (anonymous):
thanks!
OpenStudy (anonymous):
Do you need me to explain why it is extraneous?
OpenStudy (anonymous):
that would be nice
OpenStudy (anonymous):
you there or nah?
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OpenStudy (anonymous):
\[5\sqrt{x+7}=-10\]\[\sqrt{x+7}=-2\]\[x+7=4\]\[x=-3\] Now if we put x=-3 in the LHS of the original equation we get\[5\sqrt{x+7}=5\sqrt{4}\] But \[\sqrt{4}=2\ and\ NOT\ -2!\] So the LHS=10 and not -10. So the solution is extraneous.
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
Sorry, typing equations takes a long time.
OpenStudy (anonymous):
its all good thanks man!
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