Solutions to Problem Set 27

Math 101-06

11-15-2017

[Quadratic formula]

In these problems, complex solutions are okay.

1. Use the Quadratic Formula to solve

$$x^2 + 8 x - 2 = 0.$$

$$x = \dfrac{-8 \pm \sqrt{8^2 - 4(1)(-2)}}{2 \cdot 1} = \dfrac{-8 \pm \sqrt{64 + 8}}{2} = \dfrac{-8 \pm \sqrt{72}}{2} = \dfrac{-8 \pm 6 \sqrt{2}}{2} = -4 \pm 3 \sqrt{2}.\quad\halmos$$


2. Use the Quadratic Formula to solve

$$2 x^2 - 13 x + 20 = 0.$$

$$x = \dfrac{-(13) \pm \sqrt{(13)^2 - 4(2)(20)}}{2 \cdot 2} = \dfrac{13 \pm \sqrt{169 - 160}}{4} = \dfrac{13 \pm \sqrt{9}}{4} = \dfrac{13 \pm 3}{4} = 4 \quad{or}\quad \dfrac{5}{2}.\quad\halmos$$


3. Use the Quadratic Formula to solve

$$3 x^2 - 4 x - 2 = 0.$$

$$x = \dfrac{4 \pm \sqrt{(-4)^2 - 4(3)(-2)}}{2 \cdot 3} = \dfrac{4 \pm \sqrt{40}}{6} = \dfrac{4 \pm 2 \sqrt{10}}{6} = \dfrac{2 \pm \sqrt{10}}{3}.\quad\halmos$$


4. Use the Quadratic Formula to solve

$$2 x^2 - 4 x + 13 = 0.$$

$$x = \dfrac{-(-4) \pm \sqrt{(4)^2 - 4(2)(13)}}{2 \cdot 2} = \dfrac{4 \pm \sqrt{16 - 104}}{4} = \dfrac{4 \pm \sqrt{-88}}{4} = \dfrac{4 \pm 2 i \sqrt{22}}{4} = \dfrac{2 \pm i \sqrt{22}}{2}.\quad\halmos$$


5. Use the Quadratic Formula to solve

$$2 x^2 + 6 x - 11 = 0.$$

$$x = \dfrac{-6 \pm \sqrt{6^2 - 4(2)(-11)}}{2 \cdot 2} = \dfrac{-6 \pm \sqrt{36 + 88}}{4} = \dfrac{-6 \pm \sqrt{124}}{4} = \dfrac{-6 \pm 2 \sqrt{31}}{4} = \dfrac{-3 \pm \sqrt{31}}{2}.\quad\halmos$$


6. For what values of k does the following quadratic have complex roots?

$$x^2 - 6 x + 3 k = 0.$$

The Quadratic Formula contains the square root expression $\sqrt{b^2 - 4 a c}$ . If $b^2 - 4 a c$ is negative, this will be a complex number.

Hence, the quadratic $a x^2 + b
   x + c$ has complex roots when $b^2 - 4 a c < 0$ . For the given quadratic, this happens when

$$b^2 - 4 a c = (-6)^2 - 4(1)(3 k) < 0.$$

Simplify the inequality and solve:

$$\eqalign{ (-6)^2 - 4(1)(3 k) & < 0 \cr 36 - 12 k & < 0 \cr 36 & < 12 k \cr 3 & < k \cr}$$

The quadratic has complex roots for $k > 3$ .


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