Question

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.

Answer 1

maybe you know that, roots find by the formula (-b+\sqrt{b ^{2}-4ac})\div2a\]
so, b^2-4a.c must be bigger to zero for real number solutions.. i mean, 9-4*1*j bigger or equal to zero. so, \[0\le j <9/4\] for these interval, equation have 2 real solutions

Answer 2

Can you show me how to write it down on my paper and how to show my work

Answer 3

im just so confused

Answer 4

in 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 \[\sqrt{b ^{2}-4*a*c})>0\]
\[9-4*j >0\]
\[4*j<9\]
\[j<9/4\]

Answer 5

Thank you