Page 126 - aljaber

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x2 + 2 = 3x ()
x 2 − 3x + 2 = 0 ()

(x − 1)(x − 2) = 0

x −2 = 0 x −1 = 0

.x2 = 2 x1 = 1
:( )

x = 15 ± 225 − 4 × 5 × 2 = 15 ± 225 − 40
10 10

= 15 ± 185 = 3 ± 5 × 37 = 3 ± 37
10 2 100 2 20

:( )

(x + 2)(x − 1) = 2 − 3x
x2 + x − 2 = 2 − 3x

x 2 + 4x − 4 = 0

x = −4 ± 16 − 4 ×1 × (−4) = −4 ± 32 ()
22

= −4 ± 4 2 = −2 ± 2 2
2

:( )

x − 1 = 2x + 1
2−x
(2x + 1)(2 − x) = x − 1

4x − 2x 2 + 2 − x = x − 1

−2x 2 + 2x + 3 = 0

2x2 − 2x − 3 = 0