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jess.white
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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.
 one year ago
 one year ago
jess.white Group Title
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.
 one year ago
 one year ago

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raskalnikov Group TitleBest ResponseYou've already chosen the best response.1
maybe 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
 one year ago

jess.white Group TitleBest ResponseYou've already chosen the best response.0
Can you show me how to write it down on my paper and how to show my work
 one year ago

jess.white Group TitleBest ResponseYou've already chosen the best response.0
im just so confused
 one year ago

raskalnikov Group TitleBest ResponseYou've already chosen the best response.1
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 \[\sqrt{b ^{2}4*a*c})>0\] \[94*j >0\] \[4*j<9\] \[j<9/4\]
 one year ago

jess.white Group TitleBest ResponseYou've already chosen the best response.0
Thank you
 one year ago
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