Question

Determine the value of k so that 2x^2+kx+5=0 has each type of solutions:

2 real solutions:
2complex,nonreal solutions:
Exactly 1 real solution:

Answer 1

`2x^2+kx+5=0 \\ \\ a=2 \\ b=k \\ c=5 \\ \Delta=b^2-4ac=k^2-4 * 2 * 5=k^2-40`

When the discriminant Δ is greater than 0, the equation has 2 real solutions:

`k^2-40 > 0 \\ k^2>40 \\ k>√(40) \ \lor \ k<-√(40) \\ k>2√(10) \ \lor \ k<-2√(10) \\ \boxed{k \in (-\infty,-2√(10)) \cup (2√(10), +\infty)}`

When the discriminant Δ is less than 0, the equation has 2 complex and no real solutions:

`k^2-40<0 \\ k^2<40 \\ k<√(40) \ \land \ k>-√(40) \\ k<2√(10) \ \land \ k>-2√(10) \\ \boxed{k \in (-2√(10),2√(10))}`

When the discriminant Δ is equal to 0, the equation has exactly 1 real solution:

`k^2-40=0 \\ k^2=40 \\ k=√(40) \ \lor \ k=-√(40) \\ k=2√(10) \ \lor \ k=-2√(10) \\ \boxed{k \in \{ -2√(10), 2√(10) \} }`