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Write 2x^2+12x+3=0 in vertex form?

User Todmy
by
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1 Answer

4 votes

Answer:

y = 2x^2+12x+3 = 2(x + 3)^2 - 15

Explanation:

  • Take 2 as a common factor out of the first 2 terms
  • 2(x^2 + 6x) + 3 = 0
  • Add half 6 squared inside the brackets. Subtract the same number outside the brackets when multiplied by 2
  • 2(x^2 + 6x + 3^2) + 3 - 2*3^2 = 0
  • Combine the two terms on the end.
  • 2(x^2 + 6x + 9) + 3 - 2*9 = 0
  • 2(x^2 + 6x + 9) + 3 - 18 = 0
  • 2(x^2 + 6x + 9) - 15 = 0
  • Express the terms inside the brackets as a perfect square
  • 2(x + 3)^2 - 15 = 0

Discussion Graph

Technically, the vertex form would y = 2(x + 3)^2 - 15. It needs to be more general than equated to zero. I have solved it the way you presented it and graphed it by definition.

  • The red graph is y = 2x^2 + 12x + 3
  • The blue graph is y = 2(x + 3)^2 - 15

Notice they have the same critical point which means they are equivalent






Write 2x^2+12x+3=0 in vertex form?-example-1
Write 2x^2+12x+3=0 in vertex form?-example-2
User Nyegaard
by
7.6k points

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