第4章 4.1

(1)

$x^2 + 3x - 4 = 0$ $(x+4)(x-1) = 0$ $x = -4, 1$

$x = \dfrac{-3 \pm \sqrt{9+16}}{2} = \dfrac{-3 \pm 5}{2} = -4, 1$

(2)

$x^2 - 4x + 4 = 0$ $(x-2)^2 = 0$ $x = 2$

$x = 2 \pm \sqrt{4-4} = 2$

(3)

$x^2 + x - 30 = 0$ $(x+6)(x-5) = 0$ $x = -6, 5$

$x = \dfrac{-1 \pm \sqrt{1+120}}{2} = \dfrac{-1 \pm 11}{2} = -6, 5$

(4)

$2x^2 - 4x = 0$ $2x(x-2) = 0$ $x = 0, 2$

$x = \dfrac{2 \pm \sqrt{4}}{2} = \dfrac{2 \pm 2}{2} = 0, 2$

(5)

$3x^2 + 4x + 1 = 0$ $(3x+1)(x+1) = 0$ $x = -\dfrac{1}{3}, -1$

$x = \dfrac{-2 \pm \sqrt{4-3}}{3} = \dfrac{-2 \pm 1}{3} = -\dfrac{1}{3}, -1$

(6)

$2(x-1)^2 = 5(x-1) + 3$ $2(x-1)^2 - 5(x-1) - 3 = 0$

$\{2(x-1)+1\}\{(x-1)-3\} = 0$ $(2x-1)(x-4) = 0$ $x = \dfrac{1}{2}, 4$

$x - 1 = \dfrac{5 \pm \sqrt{25+24}}{4} = \dfrac{5 \pm 7}{4} = -\dfrac{1}{2}, 3$ $x = \dfrac{1}{2}, 4$