does anyone know how to do extraneous soultions for radicals?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

does anyone know how to do extraneous soultions for radicals?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

To check for extraneous solutions, plug each of your solutions in for x into the radical. If you get a negative radical, the solution was introduced through squaring and can be eliminated.
i dont understand this at all.
Do you have solutions like x=0, x=1, etc.? Take these and plug them into the original problem. If you get a negative under the radical, don't count that as an answer.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

how do you get the solutions?
\[5\sqrt{x}=1\] is one of the problems but i dont know how to find the solution
Square it, so you get 25*x=1 So x=1/25. Plug it back in 5*sqrt((1/25))=1, which works. 1/25 is not an extraneous solution.
thank you, how would i do one like this: \[(x-3)1/3=6\]
That's simple. Just multiply by 3 on both sides to get rid of the 1/3. (x-3)=6*3 x-3=18 x=21
thank you that helped me with another one similar to that, i have one last question on how to do one like this: \[\sqrt{x}+1=\sqrt{2x}-7\]
OK, let's get everything organized to be solved. 8=\[\sqrt{2x}-\sqrt{x}\] 8=\[(\sqrt{2}-1)*\sqrt{x}\] 8/\[(\sqrt{2}-1)\]=sqrt(x) square both sides. then plug in your solutions to the original equation to see if they work.

Not the answer you are looking for?

Search for more explanations.

Ask your own question