Question #704fa

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Answer 1 · 2017-10-12T23:52:28

#k=2#

For a quadratic equation to have equal roots (also known as a double root, or having a multiplicity of 2), the discriminant must equal 0.

So #b^2-4ac=0#

In this problem #a=(k-3), b=2(k-3), and c=-1#

Thus:

#(2(k-3))^2-4(k-3)(-1)=0#

#4(k^2-6k+9)+4k-12=0#

#4k^2-24k+36+4k-12=0#

#4k^2-20k+24=0#

#4(k^2-5k+6)=0#

#4(k-2)(k-3)=0#

So #k=2 or 3#

However #k=3# is an extraneous solution because if you plug in #3# into the original equation you end up getting #-1=0# which is false.

So #k=2# is the only solution.