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