How do you solve $(x + 2)^2 = 13$ ?

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Answer 1 · 2018-03-11T14:09:05

$color(red)(x = -2 + sqrt13, color(blue)( -2 - sqrt13$

$(x + 2)^2 = 13$

$x^2 + 4x + 4 - 13 = 0$

$x^2 + 4x - 9 = 0$

Standard form of quadratic equation is

$ax^2 + bx + c = 0$ and it’s roots are

$x = (-b +- sqrt (b^2 - 4ac)) / 2a$

For the given equation,

$x = (-4 +- sqrt(4^2 + (4*1*9)))/ (2 * 1)$

$x = (-4 +- sqrt(52)) / 2 = (-2 +- sqrt13)$

$color(red)(x = -2 + sqrt13, color(blue)( -2 - sqrt13$

How do you solve (x + 2)^2 = 13(x + 2)^2 = 13?