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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

@Hero

2. Diana.xL

@imqwerty

3. imqwerty

for 2 real solutions the discriminant should be a perfect square :)

4. Diana.xL

so 16?

5. imqwerty

can u tell what is the value of discriminant when u put c=16

6. Diana.xL

-48?

7. imqwerty

how

8. 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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