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

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Answer 1 · 2016-07-04T19:04:13

$x = 2 or x = 1$ $x = 2 or x = 1$

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

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

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

$x^{2} - 3 x + {(\frac{3}{2})}^{2} = - 2 + {(\frac{3}{2})}^{2}$ $x^2 - 3x + color(red)((3/2)^2) " = -2 + color(red)((3/2)^2)$

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

${(x - \frac{3}{2})}^{2} = - 2 + \frac{9}{4} = \frac{1}{4}$ $(x - 3/2)^2 = -2 + 9/4 = 1/4$

$x - \frac{3}{2} = \pm \sqrt{\frac{1}{4}}$ $x -3/2 = +-sqrt(1/4)$

$x = \pm \sqrt{\frac{1}{4}} + 1 \frac{1}{2}$ $x = +-sqrt(1/4) +1 1/2$

$x = + \frac{1}{2} + 1 \frac{1}{2} or x = - \frac{1}{2} + 1 \frac{1}{2}$ $x = +1/2 +1 1/2 " or " x = -1/2 +1 1/2$

$x = 2 or x = 1$ $x = 2 or x = 1$

How do you solve x2−3x+2=0x^2 - 3x +2 = 0 by completing the square?