For what value of #k# does the equation #4x^2 - 12x + k# have only one solution?

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Answer 1 · 2015-12-23T04:55:09

#k=9#

This quadratic function will only have one solution when the discriminant is equal to #0#.

The quadratic equation:

#x=(-b+-overbrace(sqrt(b^2-4ac))^("discriminant",Delta))/(2a)#

When the discriminant #sqrt(b^2-4ac)=0#, the only solution will be #-b/(2a)#.

In this scenario,
#a=4#
#b=-12#
#c=k#

Set the discriminant equal to zero.

#sqrt(b^2-4ac)=0#

Plug in values.

#sqrt((-12)^2-(4xx4xxk))=0#

#sqrt(144-16k)=0#

#144-16k=0#

#16k=144#

#k=9#

We can graph to check:

graph{4x^2-12x+9 [-8.5, 11.5, -2, 8]}

The graph only has one solution.

This is also provable since

#4x^2-12x+9=0#

#(2x-3)^2=0#

The only solution is #3/2#.