How do you find all the real and complex roots of #r²-14r+49=0#?

1 Answer
mason m
Jan 22, 2016

#r=7#

Explanation:

This is a perfect square trinomial—it is equal to

#(r-7)^2=0#

This can be verified by redistributing #(r-7)^2#:

#(r-7)(r-7)=r^2-7r-7r+49=r^2-14r+49#

Back to solving the equation.

#(r-7)^2=0#

Now, we can take the square root of both sides.

#r-7=0#

#r=7#

This is the only answer. Even though quadratics typically have two answers, this has a root of #7# with a multiplicity of #2#, which means that both its roots are #7#.

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