What is the value of #k# so that #kx^2-4x-3=0# has one real root?

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Answer 1 · 2017-03-25T14:40:34

#k=-4/3#

For one real root ( or equal roots) then the discriminant

#" " Delta=b^2-4ac=0#

#kx^2-4x-3=0#

#Delta=(-4)^2-4xxkxx(-3)=0#

#16+12k=0#

#k=-16/12=-4/3#

If this is correct then the resulting quadratic can be factorised as a perfect square.

#-4/3x^2-4x-3=0#

giving

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

factorising to:

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

confirming the result.