Question Number 113857 by ayenisamuel last updated on 15/Sep/20

find the values of k for which the  line y=kx−3 does not meet the curve  y=2x^2 −3x+k.

Answered by MJS_new last updated on 15/Sep/20

first, intersect them  kx−3=2x^2 −3x+k  x^2 −((k+3)/2)x+((k+3)/2)=0  let x=t+((k+3)/4) and solve for t^2   t^2 =(((k−5)(k+3))/(16))  ⇒ if (((k−5)(k+3))/(16))<0 ⇒ no solution  ⇒ −3<k<5 is the answer

Answered by 1549442205PVT last updated on 16/Sep/20

We see that  line y=kx−3 does not meet the curve  y=2x^2 −3x+k  if and only if the equation  2x^2 −3x+k=kx−3 has no roots  ⇔2x^2 −(3+k)x+k+3=0 has no roots  ⇔Δ=(3+k)^2 −4.2.(k+3)<0  ⇔k^2 −2k−15< 0⇔(k−5)(k+3)<0  ⇔−3<k<5  Answer is:        −3<k<5