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## 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.

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1. raskalnikov

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

4. raskalnikov

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

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