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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:


Sagot :

[tex]2x^2+kx+5=0 \\ \\ a=2 \\ b=k \\ c=5 \\ \Delta=b^2-4ac=k^2-4 \times 2 \times 5=k^2-40[/tex]

When the discriminant Δ is greater than 0, the equation has 2 real solutions:
[tex]k^2-40 > 0 \\ k^2>40 \\ k>\sqrt{40} \ \lor \ k<-\sqrt{40} \\ k>2\sqrt{10} \ \lor \ k<-2\sqrt{10} \\ \boxed{k \in (-\infty,-2\sqrt{10}) \cup (2\sqrt{10}, +\infty)}[/tex]

When the discriminant Δ is less than 0, the equation has 2 complex and no real solutions:
[tex]k^2-40<0 \\ k^2<40 \\ k<\sqrt{40} \ \land \ k>-\sqrt{40} \\ k<2\sqrt{10} \ \land \ k>-2\sqrt{10} \\ \boxed{k \in (-2\sqrt{10},2\sqrt{10})}[/tex]

When the discriminant Δ is equal to 0, the equation has exactly 1 real solution:
[tex]k^2-40=0 \\ k^2=40 \\ k=\sqrt{40} \ \lor \ k=-\sqrt{40} \\ k=2\sqrt{10} \ \lor \ k=-2\sqrt{10} \\ \boxed{k \in \{ -2\sqrt{10}, 2\sqrt{10} \} }[/tex]