A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
For the equation x^2+3x+j=0, find all the values of j such that the equation has two real number solutions. Show your work.
anonymous
 3 years ago
For the equation x^2+3x+j=0, find all the values of j such that the equation has two real number solutions. Show your work.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe you know that, roots find by the formula (b+\sqrt{b ^{2}4ac})\div2a\] so, b^24a.c must be bigger to zero for real number solutions.. i mean, 94*1*j bigger or equal to zero. so, \[0\le j <9/4\] for these interval, equation have 2 real solutions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you show me how to write it down on my paper and how to show my work

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in equation \[a*x ^{2}+b*x+c\] roots are \[x=(b \pm \sqrt{b ^{2}4*a*c})/2a\] this is the formula for finding roots. if \[\sqrt{b ^{2}4*a*c})=0\] equation has just one real solution. if \[\sqrt{b ^{2}4*a*c})<0\], equaiton has no real solution, if \[\sqrt{b ^{2}4*a*c})>0\] system have 2 real solution. so, if we solve \[\sqrt{b ^{2}4*a*c})>0\] \[94*j >0\] \[4*j<9\] \[j<9/4\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.