IDNLearn.com: Where curiosity meets clarity and questions find their answers. Join our knowledgeable community and get detailed, reliable answers to all your questions.
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]
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]
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.