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jess.white

  • 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.

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  1. raskalnikov
    • 3 years ago
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    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
    • 3 years ago
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    Can you show me how to write it down on my paper and how to show my work

  3. jess.white
    • 3 years ago
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    im just so confused

  4. raskalnikov
    • 3 years ago
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    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
    • 3 years ago
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    Thank you

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