Solutions to Problem Set 29

Math 101-06

11-20-2017

[Quadratic in form]

1. Solve the equation for x. Complex numbers are allowed.

$$(3 x - 1)^2 - 4(3 x - 1) - 5 = 0.$$

Don't multiply out!

Let $y = 3 x - 1$ . Then

$$\matrix{ & & y^2 - 4 y - 5 = 0 & & \cr & & (y - 5)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y - 5 = 0 & & & & y + 1 = 0 \cr y = 5 & & & & y = -1 \cr 3 x - 1 = 5 & & & & 3 x - 1 = -1 \cr 3 x = 6 & & & & 3 x = 0 \cr x = 2 & & & & x = 0 \cr}$$

The solutions are $x = 2$ and $x = 0$ .


2. Solve the equation for x. Complex numbers are allowed.

$$x^6 - 7 x^3 - 8 = 0.$$

Since $x^6 = (x^3)^2$ , the equation can be written this way:

$$(x^3)^2 - 7 x^3 - 8 = 0.$$

Let $y = x^3$ . Then

$$\matrix{ & & y^2 - 7 y - 8 = 0 & & \cr & & (y - 8)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y - 8 = 0 & & & & y + 1 = 0 \cr y = 8 & & & & y = -1 \cr x^3 = 8 & & & & x^3 = -1 \cr x = 2 & & & & x = -1 \cr}$$

The solutions are $x = 2$ and $x = -1$ .


3. Solve the equation for x. Complex numbers are allowed.

$$x^4 - 6 x^3 + 5 = 0.$$

Since $x^4 = (x^2)^2$ , the equation can be written this way:

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

Let $y = x^2$ . Then

$$\matrix{ & & y^2 - 6 y + 5 = 0 & & \cr & & (y - 5)(y - 1) = 0 & & \cr & \swarrow & & \searrow & \cr y - 5 = 0 & & & & y - 1 = 0 \cr y = 5 & & & & y = 1 \cr x^2 = 5 & & & & x^2 = 1 \cr x = \pm \sqrt{5} & & & & x = \pm 1 \cr}$$

The solutions are $x = \pm
   \sqrt{5}$ and $x = \pm 1$ .


4. Solve the equation for x. Complex numbers are allowed.

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

Since $x^4 = (x^2)^2$ , the equation can be written this way:

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

Let $y = x^2$ . Then

$$\matrix{ & & y^2 + 3 y - 4 = 0 & & \cr & & (y + 4)(y - 1) = 0 & & \cr & \swarrow & & \searrow & \cr y + 4 = 0 & & & & y - 1 = 0 \cr y = -4 & & & & y = 1 \cr x^2 = -4 & & & & x^2 = 1 \cr x = \pm 2 i & & & & x = \pm 1 \cr}$$

The solutions are $x = \pm 2 i$ and $x = \pm 1$ .


5. Solve the equation for x. Complex numbers are allowed.

$$x - 5 \sqrt{x} - 6 = 0.$$

Since $x = (\sqrt{x})^2$ , the equation can be written this way:

$$(\sqrt{x})^2 - 5 \sqrt{x} - 6 = 0.$$

Let $y = \sqrt{x}$ . Then

$$\matrix{ & & y^2 - 5 y - 6 = 0 & & \cr & & (y + 1)(y - 6) = 0 & & \cr & \swarrow & & \searrow & \cr y + 1 = 0 & & & & y - 6 = 0 \cr y = -1 & & & & y = 6 \cr \sqrt{x} = -1 & & & & \sqrt{x} = 6 \cr}$$

$\sqrt{x} = -1$ has no solutions, because $\sqrt{x}$ can't be negative.

$\sqrt{x} = 6$ gives

$$(\sqrt{x})2 = 6^2, \quad\hbox{so}\quad x = 36.$$

The solution is $x = 36$ .


6. Solve the equation for x. Complex numbers are allowed.

$$\dfrac{1}{x^2} - \dfrac{2}{x} - 15 = 0.$$

Let $y = \dfrac{1}{x}$ . Then

$$\matrix{ & & y^2 - 2 y - 15 = 0 & & \cr & & (y - 5)(y + 3) = 0 & & \cr & \swarrow & & \searrow & \cr y - 5 = 0 & & & & y + 3 = 0 \cr y = 5 & & & & y = -3 \cr \noalign{\vskip3pt} \dfrac{1}{x} = 5 & & & & \dfrac{1}{x} = -3 \cr \noalign{\vskip3pt} x \cdot \dfrac{1}{x} = x \cdot 5 & & & & x \cdot \dfrac{1}{x} = x \cdot (-3) \cr \noalign{\vskip3pt} 1 = 5 x & & & & 1 = -3 x \cr \noalign{\vskip3pt} \dfrac{1}{5} = x & & & & -\dfrac{1}{3} = x \cr}$$

The solutions are $x =
   \dfrac{1}{5}$ and $x = -\dfrac{1}{3}$ .


One hears only those questions for which one is able to find answers. - Friedrich Nietzsche


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