What factoring pattern can you use to solve x^2+14x+49=0

A (ax+√c)^2=0

B (√a*x+√c)^2=0

C (√a*x+c)(√a*x-c)=0

D (ax+√c)(ax-√c)=c

Answers

Answer 1

Answer:

  B  (√a*x+√c)^2=0

Step-by-step explanation:

It can be helpful to look a the expanded form of each of these factoring patterns.

  A (ax+√c)^2=0   ⇒   a^2x^2 +2a√c*x +c = 0

  B (√a*x+√c)^2=0   ⇒   ax^2 +2√(ac)*x +c = 0

  C (√a*x+c)(√a*x-c)=0   ⇒   ax^2 -c^2 = 0

  D (ax+√c)(ax-√c)=c   ⇒   a^2x^2 -c = c

If 'a' and 'c' represent the coefficients in the given equation, then choices A, C, and D can be eliminated. The expanded forms of those end up with a^2 and/or c^2 as coefficients, not 'a' and 'c'.

Choice B is the only sensible one, and fits the given equation.

  a = 1, c = 49   ⇒   √a = 1, √c = 7

  x^2 +14x +49 = (x +7)^2 = 0 . . . . . factored using Pattern B


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