Diana.xL
  • Diana.xL
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Diana.xL
  • Diana.xL
@Hero
Diana.xL
  • Diana.xL
@imqwerty
imqwerty
  • imqwerty
for 2 real solutions the discriminant should be a perfect square :)

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More answers

Diana.xL
  • Diana.xL
so 16?
imqwerty
  • imqwerty
can u tell what is the value of discriminant when u put c=16
Diana.xL
  • Diana.xL
-48?
imqwerty
  • imqwerty
how
campbell_st
  • campbell_st
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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