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Question Number 120035 by I want to learn more last updated on 28/Oct/20

Answered by ebi last updated on 29/Oct/20

y=2x−12......(1) y=px^2 −4px−5p......(2) Substitute (1) into (2) 2x−12=px^2 −4px−5p px^2 −2(2p+1)x−(5p−12)=0......(3) a=p b=−2(2p+1) c=−(5p−12) (a) Since (3) doesnt have real roots, the discriminant <0 b^2 −4ac<0 [−2(2p+1)]^2 −4(p)[−(5p−12)]<0 4(4p^2 +4p+1)+4p(5p−12)]<0 16p^2 +16p+4+20p^2 −48p<0 36p^2 −32p+4<0 9p^2 −8p+1<0 (shown) (b) 9p^2 −8p+1<0 p^2 −(8/9)p<−(1/9) p^2 −(8/9)p+(−(4/9))^2 <−(1/9)+(−(4/9))^2 (p−(4/9))^2 <(7/(81)) p−(4/9)<((√7)/9) or p−(4/9)>−((√7)/9) p<((4+(√7))/9) or p>((4−(√7))/9) p<0.7384 or p>0.1505 ∴the range is 0.1505<p<0.7384

y=2x−12......(1)y=px2−4px−5p......(2)Substitute(1)into(2)2x−12=px2−4px−5ppx2−2(2p+1)x−(5p−12)=0......(3)a=pb=−2(2p+1)c=−(5p−12)(a)Since(3)doesnthaverealroots,thediscriminant<0b2−4ac<0[−2(2p+1)]2−4(p)[−(5p−12)]<04(4p2+4p+1)+4p(5p−12)]<016p2+16p+4+20p2−48p<036p2−32p+4<09p2−8p+1<0(shown)(b)9p2−8p+1<0p2−89p<−19p2−89p+(−49)2<−19+(−49)2(p−49)2<781p−49<79orp−49>−79p<4+79orp>4−79p<0.7384orp>0.1505∴therangeis0.1505<p<0.7384

Commented by I want to learn more last updated on 29/Oct/20

Thanks sir

Thankssir

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