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](Q120061.png)
$${y}=\mathrm{2}{x}−\mathrm{12}......\left(\mathrm{1}\right) \\ $$$${y}={px}^{\mathrm{2}} −\mathrm{4}{px}−\mathrm{5}{p}......\left(\mathrm{2}\right) \\ $$$${Substitute}\:\left(\mathrm{1}\right)\:{into}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{x}−\mathrm{12}={px}^{\mathrm{2}} −\mathrm{4}{px}−\mathrm{5}{p} \\ $$$${px}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{p}+\mathrm{1}\right){x}−\left(\mathrm{5}{p}−\mathrm{12}\right)=\mathrm{0}......\left(\mathrm{3}\right) \\ $$$${a}={p}\:\:\:\:{b}=−\mathrm{2}\left(\mathrm{2}{p}+\mathrm{1}\right)\:\:\:\:{c}=−\left(\mathrm{5}{p}−\mathrm{12}\right) \\ $$$$\left({a}\right) \\ $$$${Since}\:\left(\mathrm{3}\right)\:{doesnt}\:{have}\:{real}\:{roots}, \\ $$$${the}\:{discriminant}\:<\mathrm{0} \\ $$$${b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0} \\ $$$$\left[−\mathrm{2}\left(\mathrm{2}{p}+\mathrm{1}\right)\right]^{\mathrm{2}} −\mathrm{4}\left({p}\right)\left[−\left(\mathrm{5}{p}−\mathrm{12}\right)\right]<\mathrm{0} \\ $$$$\left.\mathrm{4}\left(\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}{p}+\mathrm{1}\right)+\mathrm{4}{p}\left(\mathrm{5}{p}−\mathrm{12}\right)\right]<\mathrm{0} \\ $$$$\mathrm{16}{p}^{\mathrm{2}} +\mathrm{16}{p}+\mathrm{4}+\mathrm{20}{p}^{\mathrm{2}} −\mathrm{48}{p}<\mathrm{0} \\ $$$$\mathrm{36}{p}^{\mathrm{2}} −\mathrm{32}{p}+\mathrm{4}<\mathrm{0} \\ $$$$\mathrm{9}{p}^{\mathrm{2}} −\mathrm{8}{p}+\mathrm{1}<\mathrm{0}\:\left({shown}\right) \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\mathrm{9}{p}^{\mathrm{2}} −\mathrm{8}{p}+\mathrm{1}<\mathrm{0} \\ $$$${p}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{9}}{p}<−\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${p}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{9}}{p}+\left(−\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{2}} <−\frac{\mathrm{1}}{\mathrm{9}}+\left(−\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\left({p}−\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{2}} <\frac{\mathrm{7}}{\mathrm{81}} \\ $$$$ \\ $$$${p}−\frac{\mathrm{4}}{\mathrm{9}}<\frac{\sqrt{\mathrm{7}}}{\mathrm{9}}\:\:\:{or}\:\:\:\:\:{p}−\frac{\mathrm{4}}{\mathrm{9}}>−\frac{\sqrt{\mathrm{7}}}{\mathrm{9}} \\ $$$${p}<\frac{\mathrm{4}+\sqrt{\mathrm{7}}}{\mathrm{9}}\:\:\:\:\:{or}\:\:\:\:\:\:{p}>\frac{\mathrm{4}−\sqrt{\mathrm{7}}}{\mathrm{9}} \\ $$$${p}<\mathrm{0}.\mathrm{7384}\:\:{or}\:\:\:\:\:{p}>\mathrm{0}.\mathrm{1505} \\ $$$$ \\ $$$$\therefore{the}\:{range}\:{is}\:\mathrm{0}.\mathrm{1505}<{p}<\mathrm{0}.\mathrm{7384} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by I want to learn more last updated on 29/Oct/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$