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Diana.xL

  • one year ago

If there are two real solutions to the equation x^2 + 4x + c=0 , which is a possible value of c? 20 17 -5 16

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  1. Diana.xL
    • one year ago
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    @Hero

  2. Diana.xL
    • one year ago
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    @imqwerty

  3. imqwerty
    • one year ago
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    for 2 real solutions the discriminant should be a perfect square :)

  4. Diana.xL
    • one year ago
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    so 16?

  5. imqwerty
    • one year ago
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    can u tell what is the value of discriminant when u put c=16

  6. Diana.xL
    • one year ago
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    -48?

  7. imqwerty
    • one year ago
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    how

  8. campbell_st
    • one year ago
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    well to have 2 real solution, then the discriminant is \[b^2 - 4ac \ge 0\] and the result is either 0 or a square number... so you have a = 1, b = 4 and c then \[4^2- 4 \times 1 \times c \ge 0\] then \[16 - 4c \ge 0 \] so solve the inequality then find the solution that meets the condition of the inequality. don't forget, dividing by a negative requires the inequality to be reversed. and just a quick revision on the discriminant \[b^2 - 4ac = 0\] there are 2 equal roots \[b^2 - 4ac >0 ~~and~~b^2 - 4ac ~~is~a~\square ~number\] then the roots, are real, unequal and rational. Hope it makes sense

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