How do you show that #kx^2+2x-(k-2) = 0# has at least one real root for any value of #k# ?

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Answer 1 · 2015-05-29T20:02:53

#kx^2+2x-(k-2)# has discriminant #Delta# given by the formula

#Delta = 2^2-(4k*-(k-2))#

#=4+4k^2-8k#

#=4(k^2-2k+1)#

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

If #k=1# then #Delta = 0# and #kx^2+2x-(k-2) = 0# has one repeated real root #(x = -1)#.

If #k != 1# then #Delta > 0# and #kx^2+2x-(k-2) = 0# has 2 distinct real roots.

Answer 2 · 2015-05-29T21:03:35

More simply, #x = -1# is a root for all values of #k#

Substituting #x=-1#, we get:

#kx^2+2x-(k-2) = k(-1)^2+2(-1)-(k-2)#

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

So #(x+1)# is one factor of #kx^2+2x-(k-2)#, the other being #(kx-k+2)#

#(x+1)(kx-k+2)#

#=x(kx-k+2)+(kx-k+2)#

#=kx^2-kx+2x+kx-k+2#

#=kx^2+2x-(k-2)#

The other root of #kx^2+2x-(k-2) = 0# is #x = (k-2)/k# unless #k=0#.