## jess.white 2 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.

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

2. jess.white

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

3. jess.white

im just so confused

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

5. jess.white

Thank you