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A boggart is a shape shifter. It lives in dark places and when confronted, changes into whatever the person fears most.
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What Harry Potter creature is unforgiving?
The Boggart.
What is Seamus's boggart?
His boggart is a banshee
When was The Boggart created?
The Boggart was created in 1993.
What is Molly Weasleys boggart?
Molly Weasley's Boggart is her Dead family/friends. Her Boggart is mentioned in the novels, specifically, in 'Harry Potter and The Order of the Phoenix' I believe her Boggart was omitted from the movie.
What is George and Fred's boggart?
In the 3rd book they do not have Fred and george in front of a boggart Ron's was spiders, Neville's was professor snape, harrys would be voldemort than he thought of the dementors, molly weasley's was everyone dead that she loved, lupins was the full moon
What is seamus?
His boggart is a banshee
What is Hermiones Boggart?
A boggart is a shape-shifting creature that takes on the form of its intended victim's worst fear. It generally likes to hide in dark, enclosed places.In Harry Potter and the Prisoner of Azkaban, Professor Remus Lupin teaches his students in Defence Against the Dark Arts to approach a boggart in groups of two or more, so that the boggart will have difficulty in choosing which one to frighten. A common wizard's defense against a boggart is to point a wand at the boggart and say "Riddikulus" while thinking of something very funny; this charm puts out a protective shield so that the boggart can be held at bay, or manuevered away.
What is the name of the two mice in spiderwick books?
The two mice in the Spiderwick Chronicles are named Byron and Boggart. Byron is a common house mouse, while Boggart is a newly discovered species called a "kumis."
What is it called when the basic problem is solved in a book?
It is the Solution
Who is Lord Voldemort's boggart in LEGO Harry Potter?
Lord Voldemort's boggart is his own dead body. His riddikulus is Harry drooping dead.
I am not sure you've had enough coffee for this?
Boggart.
Is Dean Thomas' boggart a Banshee?
Yes.
