How do you solve $x^2 - 3x +2 = 0$ by completing the square?

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Answer 1 · 2016-07-04T19:04:13

$x = 2 or x = 1$

$ax^2 + bx +c =0$ is the general form of a quadratic trinomial.

$x^2 - 3x " " = -2 " move the constant to the RHS"$

Add $color(red)((b/2)^2)$ to both sides to form the square of a binomial. This step is the completing of the square.

$x^2 - 3x + color(red)((3/2)^2) " = -2 + color(red)((3/2)^2)$

Add $(b/2)^2$ to both sides to form the square of a binomial.

$(x - 3/2)^2 = -2 + 9/4 = 1/4$

$x -3/2 = +-sqrt(1/4)$

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

$x = +1/2 +1 1/2 " or " x = -1/2 +1 1/2$

$x = 2 or x = 1$

How do you solve x^2 - 3x +2 = 0x^2 - 3x +2 = 0 by completing the square?