How do you solve $4 x^{2} + 4 x + 1 = 49$ $4x^2 + 4x + 1 = 49$ by completing the square?

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Answer 1 · 2015-05-23T04:12:01

y = 4x^2 + 4x + 1 = 49

${(2 x + 1)}^{2} = 49$ $(2x + 1)^2 = 49$

$(2 x + 1) = \pm 7$ $(2x + 1) = +-7$ -> $2 x = - 1 \pm 7 \to x = - \frac{1}{2} \pm \frac{7}{2}$ $2x = -1 +- 7 -> x = -1/2 +- 7/2$

How do you solve 4x^2 + 4x + 1 = 494x2+4x+1=494x^2 + 4x + 1 = 49 by completing the square?