How do you solve $x^{2} + 4 x = 7$ $x^2 + 4x = 7$ by completing the square?

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Answer 1 · 2017-02-27T11:55:10

$x = \pm \sqrt{11} - 2$ $x=+-sqrt11-2$

$x^{2} + 4 x = 7$ $x^2+4x=7$ ; Add $4$ $4$ to both sides,

$x^{2} + 4 x + 4 = 11$ $x^2+4x+4=11$ ; now the LHS is a perfect square

$(x + 2) (x + 2) = 11$ $(x+2)(x+2)=11$

${(x + 2)}^{2} = 11$ $(x+2)^2=11$

Take the square root of both sides........

$x + 2 = \pm \sqrt{11}$ $x+2=+-sqrt(11)$

$x = \pm \sqrt{11} - 2$ $x=+-sqrt11-2$

How do you solve x2+4x=7x^2 + 4x = 7 by completing the square?