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anonymous
 one year ago
Solve 2x2 + 5x + 5 = 0. Round solutions to the nearest hundredth.
No real solutions
x ≈ 3.27 and x ≈ 0.77
x ≈ 2.98 and x ≈ 7.02
x ≈ 0.77 and x ≈ 3.27
anonymous
 one year ago
Solve 2x2 + 5x + 5 = 0. Round solutions to the nearest hundredth. No real solutions x ≈ 3.27 and x ≈ 0.77 x ≈ 2.98 and x ≈ 7.02 x ≈ 0.77 and x ≈ 3.27

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campbell_st
 one year ago
Best ResponseYou've already chosen the best response.3you can use the determinant... to check the solutions or use the general quadratic formula to solve this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so a=2, b=5, and c=5, yeah?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@campbell_st is it A, because i got x=0.43, and x=2.07 and those arent options.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.3well you shouldn't have been able to get answers... as the discriminant in this question is \[b^2  4ac = 5^2  4 \times 2 \times 5 = 15\] and you can't take the square root of a negative number... so you choice of A is correct. but your method is flawed
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