How do you solve ${(x + 1)}^{2} - 3 = 13$ $(x+1)^2-3=13$ ?

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Answer 1 · 2016-09-06T17:15:15

$x = 3 or x = - 5$ $x = 3 or x = -5$

Although this is a quadratic equation, we do not have to use the normal method of making it equal to 0. This is a special case - there is no x-term.

Move all the constants to the right hand side and then find the square root of each side.

${(x + 1)}^{2} = 16$ $(x+1)^2 = 16$

$x + 1 = \pm \sqrt{16} = \pm 4$ $x+1 = +-sqrt16 = +-4$

$if x+1 =+4 " " rarr x = 3$

$if x+1 = -4" " rarr x = -5$

How do you solve (x+1)^2-3=13(x+1)2−3=13 (x+1)^2-3=13?