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Solve 3x^2 + 2x + 7 = 0. Round soultions to nearest hundreth

Solve 3x^2 + 2x + 7 = 0. Round soultions to nearest hundreth-example-1
User Kevmon
by
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1 Answer

5 votes

Answer:

No real solutions

Explanation:

  • 3x^2 + 2x + 7 = 0

Solution 1

From the first look, the equation has to have two positive roots as all coefficients (a,b,c) are positive

Since the are no two positive solutions given as an option, the last choice is correct

Solution 2

  • 3x^2 + 2x + 7 = 0
  • x^2 + 2x + 1 + 2x^2 + 6 = 0
  • (x + 1)^2 + 2x^2 + 6 = 0

As we see all terms are positive, therefore there are no real solutions

Solution 3

  • D = b^2 - 4ac = 2^2 - 4*3*7 = 4 - 84 = -80

Since discriminant is negative, there are no real solutions

User Elliott Davies
by
8.5k points

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