Find the set of values of k for which the equation #x^2#+2(k-1)x+k-1=0 has no real roots. ?

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Answer 1 · 2018-07-02T14:47:16

The solution is #k in (1,2)#

The quadratic equation is

#x^2+2(k-1)x+(k-1)=0#

In order for the equations to have no real roots, the discriminant must be #<0#

Therefore,

#Delta=b^2-4ac=(2(k-1))^2-4(1)(k-1)#

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

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

#=4k^2-12k+8#

So,

#4k^2-12k+8<0#

#=>#, #k^2-3k+2<0#

#=>#, #(k-2)(k-1)<0#

The solutions to this inequality is #k in (1,2)#

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