Question 4.2: Solve each of the following equations, classifying the type ......

Use equation (4.2), the ‘minus b’ formula, to solve for the values of x.

(a)

x² + 6x + 5 =0

x = x=\frac{-(b) \pm \sqrt{(b)^{2}  –  4(a)(c)}}{2(a)}

a = 1, b = 6, c = 5

x = \frac{-6  \pm \sqrt{(6)^{2}  –  4(1)(5)}}{2(1)}

x = \frac{-6  \pm \sqrt{36  −  20}}{2}

x = \frac{-6  \pm \sqrt{16}}{2}

x = \frac{-6  \pm  4}{2}

root 1 = \frac{−6  + 4}{2} = \frac{−2}{2} = −1

root 2 = \frac{−6   – 4}{2} = \frac{−10}{2} = −5

Different solutions (real roots)

 (b)

x² + 6x + 9 =0

x = x=\frac{-(b) \pm \sqrt{(b)^{2}  –  4(a)(c)}}{2(a)}

a = 1, b = 6, c = 9

x = \frac{-6  \pm \sqrt{(6)^{2}  –  4(1)(9)}}{2(1)}

x = \frac{-6  \pm \sqrt{36  −  36}}{2}

x = \frac{-6  \pm \sqrt{0}}{2}

x = \frac{-6  \pm  0}{2}

root 1 = \frac{−6  + 0}{2} = \frac{−6}{2} = −3

root 2 = \frac{−6   – 0}{2} = \frac{−6}{2} = −3

Repeated solutions (real roots)

 (c)

x² + 6x + 10 = 0

x = x=\frac{-(b) \pm \sqrt{(b)^{2}  –  4(a)(c)}}{2(a)}

a = 1, b = 6, c = 10

x = \frac{-6  \pm \sqrt{(6)^{2}  –  4(1)(10)}}{2(1)}

x = \frac{-6  \pm \sqrt{36  −  40}}{2}

x = \frac{-6  \pm \sqrt{-4}}{2}

x = \frac{-6  \pm  2i}{2}

root 1 = \frac{−6  + 2i}{2} = =−3 + i

root 2 = \frac{−6   – 2i}{2} = = −3 + i

Different solutions (complex roots)