Answer:
(i) [tex]k\in (-\infty,-16)\cup (0,\infty)[/tex]
(ii) [tex]p\in (9,\infty)[/tex]
Step-by-step explanation:
(i) Given equation
[tex]2x^2+4x+2=k(x+3)[/tex]
Rewrite this equation as quadratic equation in the standard form:
[tex]2x^2+4x+2=kx+3k\\ \\2x^2+4x+2-kx-3k=0\\ \\2x^2+(4-k)x+(2-3k)=0[/tex]
If the equation has two distinct real roots, then its discriminant is greater than 0. Find the discriminant:
[tex]D=(4-k)^2-4\cdot 2\cdot (2-3k)=16-8k+k^2-16+24k=k^2+16k[/tex]
Solve the inequality [tex]D>0:[/tex]
[tex]k^2+16k>0\\ \\k(k+16)>0\Rightarrow k\in (-\infty,-16)\cup (0,\infty)[/tex]
(ii) The quadratic expression is always greater than 0 when:
[tex]p>0\\ \\D<0[/tex]
Find the discriminant:
[tex]D=6^2-4\cdot p\cdot 1=36-4p[/tex]
Solve the inequality [tex]D<0:[/tex]
[tex]36-4p<0\\ \\-4p<-36\\ \\4p>36\\ \\p>9[/tex]
Assuming that [tex]p>0[/tex] and [tex]p>9,[/tex] you get
[tex]p>9\Rightarrow p\in (9,\infty)[/tex]
