Review Sheet 4

Math 101

8-6-2012

These problems are on material we'll cover after Test 3. You can use it and the Final Review Sheet to study for the final.

1. Solve the following quadratic equations:

(a) $(x - 4)^2 = 25$ .

(b) $(2x - 3)^2 = -16$ .

(c) $x^2 + 4x + 1 = 0$ .

(d) $x^2 - 6x + 25 = 0$ .

(e) $10x^2 + 18 = 27x$ .

(f) $x^2 - 4x + 29 = 0$ .

(g) $x^2 + 2x = -10$ .

2. Given the value of $b^2 - 4ac$ for a quadratic equation $ax^2 + bx + c = 0$ , tell what kind of roots the equation has.

(a) $b^2 - 4ac = 31$ .

(b) $b^2 - 4ac = 0$ .

(c) $b^2 - 4ac = -15$ .

3. (a) Show that no matter what k is, the following equation has complex roots:

$$x^2 - 4x + (k^2 + 5) = 0.$$

(b) For what value or values of p does the equation $2x^2 - px + 50 = 0$ have exactly one root?

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

5. Solve $x^4 + 2 x^2 - 8 = 0$ for x.

6. Solve $x^{-2} - 4 x^{-1} + 3 = 0$ for x.

7. Find the distance from $(3,-4)$ to $(7,1)$ .

8. Find the center and radius of the circle

$$x^2 + 6x + y^2 - 14y = 6.$$

9. Find the center and radius of the circle

$$x^2 + 3x + y^2 - 4y = 6.$$

10. Graph the parabola $y = 3 + 2 x - x^2$ . Find the roots and the x and y-coordinates of the vertex.

11. Graph the parabola $y = x^2 - 4 x + 5$ . Find the roots and the x and y-coordinates of the vertex.

12. The area of a rectangle is 84 square miles. The length is 4 miles less than 3 times the width. Find the dimensions.

13. The length of a rectangle is 2 less than 3 times the width. The area is 176. Find the dimensions of the rectangle.

14. The sum of two numbers is 5. The sum of their reciprocals is $\dfrac{45}{44}$ . Find the two numbers.

15. Calvin and Bonzo, eating together, can eat 540 rib sandwiches in 6 hours. Eating alone, Calvin can eat 240 rib sandwiches in 4 hours less than it takes Bonzo, eating alone, to eat 240 rib sandwiches. How long does it take Calvin, eating alone, to eat 240 rib sandwiches?

16. A river flows at the rate of 8 feet per second. It takes Calvin 2 seconds longer to ride his boat 384 feet upstream against the current than it takes him to ride his boat 384 feet downstream with the current. Find the speed of Calvin's boat in still water (in feet per second).

17. Solve the inequality $x^2(x - 3)(x + 8) < 0$ .

18. Solve the inequality $\dfrac{x^2}{x^2 - x - 6} > 0$ .

Solutions to Review Sheet 4

1. Solve the following quadratic equations:

(a) $(x - 4)^2 = 25$ .

$$\eqalign{(x - 4)^2 &= 25 \cr x - 4 &= \pm 5 \cr}$$

$x - 4 = 5$ gives $x = 9$ . $x - 4 = -5$ gives $x = -1$ .

The solutions are $x = 9$ and $x = -1$ .

(b) $(2x - 3)^2 = -16$ .

$$\eqalign{(2x - 3)^2 &= -16 \cr 2x - 3 &= \pm 4i \cr}$$

(I simplified $\sqrt{-16}$ by $\sqrt{-16} = \sqrt{16}\sqrt{-1} = 4i$ .)

$2x - 3 = 4i$ gives $2x = 3 + 4i$ , or $x = \dfrac{3 + 4i}{2}$ . $2x - 3 = -4i$ gives $2x = 3 - 4i$ , or $x = \dfrac{3 - 4i}{2}$ .

The solutions are $x = \dfrac{3 \pm 4i}{2}$ .

(c) $x^2 + 4x + 1 = 0$ .

Use the quadratic formula:

$$x = \dfrac{-4 \pm \sqrt{16 - 4}}{2} = \dfrac{-4 \pm \sqrt{12}}{2} = \dfrac{-4 \pm \sqrt{4}\sqrt{3}}{2} = \dfrac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}.$$

The roots are $x = -2 + \sqrt{3}$ and $x = -2 - \sqrt{3}$ .

(d) $x^2 - 6x + 25 = 0$ .

Use the quadratic formula:

$$x = \dfrac{6 \pm \sqrt{36 - 100}}{2} = \dfrac{6 \pm \sqrt{-64}}{2} = \dfrac{6 \pm \sqrt{64}\sqrt{-1}}{2} = \dfrac{6 \pm 8i}{2} = 3 \pm 4i.$$

The roots are $x = 3 \pm 4i$ .

(e) $10x^2 + 18 = 27x$ .

$$\eqalign{10x^2 + 18 &= 27x \cr 10x^2 - 27x + 18 &= 0 \cr}$$

Apply the quadratic formula:

$$x = \dfrac{27 \pm \sqrt{729 - 720}}{20} = \dfrac{27 \pm \sqrt{9}}{20} = \dfrac{27 \pm 3}{20}.$$

$x = \dfrac{27 + 3}{20} = \dfrac{3}{2}$ and $x = \dfrac{27 - 3}{20} = \dfrac{6}{5}$ . The solutions are $x = \dfrac{3}{2}$ and $x = \dfrac{6}{5}$ .

(f) $x^2 - 4x + 29 = 0$ .

Apply the quadratic formula:

$$x = \dfrac{4 \pm \sqrt{16 - 116}}{2} = \dfrac{4 \pm \sqrt{-100}}{2} = \dfrac{4 \pm \sqrt{100}\sqrt{-1}}{2} = \dfrac{4 \pm 10i}{2} = 2 \pm 5i.\quad\halmos$$

(g) $x^2 + 2x = -10$ .

$$\eqalign{x^2 + 2x &= -10 \cr x^2 + 2x + 10 &= 0 \cr}$$

Apply the quadratic formula:

$$x = \dfrac{-2 \pm \sqrt{4 - 40}}{2} = \dfrac{-2 \pm \sqrt{-36}}{2} = \dfrac{-2 \pm \sqrt{36}\sqrt{-1}}{2} = \dfrac{-2 \pm 6i}{2} = -1 \pm 3i.\quad\halmos$$

2. Given the value of $b^2 - 4ac$ for a quadratic equation $ax^2 + bx + c = 0$ , tell what kind of roots the equation has.

(a) $b^2 - 4ac = 31$ .

$b^2 - 4ac$ is a positive number, so there are two (different) real roots.

(b) $b^2 - 4ac = 0$ .

$b^2 - 4ac$ is zero, so there is one (double) real root.

Note: This happens when the equation is something like "$(x - 7)^2 = 0$ ". The only root is $x = 7$ , but it's a "double" root because the factor of $x - 7$ appears twice (squared).

(c) $b^2 - 4ac = -15$ .

$b^2 - 4ac$ is a negative number, so there are two complex roots.

3. (a) Show that no matter what k is, the following equation has complex roots:

$$x^2 - 4x + (k^2 + 5) = 0.$$

The discriminant is

$$b^2 - 4ac = 16 - 4(k^2 + 5) = 16 - 4k^2 - 20 = -4 - 4k^2.$$

Since $k^2$ is nonegative, $-4k^2$ is always less than or equal to 0. Therefore, $-4 - 4k^2$ is negative. Since the discriminant is negative no matter what k is, the equation always has complex roots.

(b) For what value or values of p does the equation $2x^2 - px + 50 = 0$ have exactly one root?

The discriminant is

$$b^2 - 4ac = p^2 - 400.$$

The equation has exactly one root when the discriminant is 0:

$$\eqalign{p^2 - 400 &= 0 \cr p^2 &= 400 \cr p &= \pm 20 \cr}$$

The equation has exactly one root when $p = 20$ or $p = -20$ .

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

Write the given equation as

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

Let $y = x^3$ . Then

$$\matrix{ & & y^2 - 5 y + 4 = 0 & & \cr & & (y - 1)(y - 4) = 0 & & \cr & \searrow & & \searrow & \cr y = 1 & & & & y = 4 \cr x^3 = 1 & & & & x^3 = 4 \cr x = 1 & & & & x = \root 3 \of 4 \quad\halmos ]cr}$$

5. Solve $x^4 + 2 x^2 - 8 = 0$ for x.

Write the given equation as

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

Let $y = x^2$ . Then

$$\matrix{ & & y^2 + 2 y - 8 = 0 & & \cr & & (y + 4)(y - 2) = 0 & & \cr & \searrow & & \searrow & \cr y = -4 & & & & y = 2 \cr x^2 = -4 & & & & x^2 = 2 \cr x = \pm 2 i & & & & x = \pm \sqrt{2} \quad\halmos ]cr}$$

6. Solve $x^{-2} - 4 x^{-1} + 3 = 0$ for x.

Write the equation as

$$(x^{-1})^2 - 4 x^{-1} + 3 = 0.$$

Let $y = x^{-1}$ . Then

$$\matrix{ & & y^2 - 4 y + 3 = 0 & & \cr & & (y - 3)(y - 1) = 0 & & \cr & \swarrow & & \searrow & \cr y = 3 & & & & y = 1 \cr \noalign{\vskip2pt} x^{-1} = 3 & & & & x^{-1} = 1 \cr \noalign{\vskip2pt} \dfrac{1}{x} = 3 & & & & \dfrac{1}{x} = 1 \cr \noalign{\vskip2pt} x = \dfrac{1}{3} & & & & x = 1 \quad\halmos \cr}$$

7. Find the distance from $(3,-4)$ to $(7,1)$ .

$$\hbox{distance} = \sqrt{(3 - 7)^2 + (-4 - 1)^2} = \sqrt{16 + 25} = \sqrt{41}.\quad\halmos$$

8. Find the center and radius of the circle

$$x^2 + 6x + y^2 - 14y = 6.$$

To complete the square in x, I need to add $\left(\dfrac{6}{2}\right)^2 = 3^2 = 9$ .

To complete the square in y, I need to add $\left(\dfrac{-14}{2}\right)^2 = (-7)^2 = 49$ .

So I get

$$\eqalign{x^2 + 6x + 9 + y^2 - 14y + 49 &= 6 + 9 + 49 \cr (x + 3)^2 + (y - 7)^2 &= 64 \cr}$$

The center is $(-3,7)$ and the radius is $\sqrt{64} = 8$ .

9. Find the center and radius of the circle

$$x^2 + 3x + y^2 - 4y = 6.$$

To complete the square in x, I need to add $\left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$ .

To complete the square in y, I need to add $\left(\dfrac{-4}{2}\right)^2 = (-)^2 = 4$ .

So I get

$$\eqalign{x^2 + 3x + \dfrac{9}{4} + y^2 - 4y + 4 &= 6 + \dfrac{9}{4} + 4 \cr \left(x + \dfrac{3}{2}\right)^2 + (y - 2)^2 &= \dfrac{49}{4} \cr}$$

The center is $\left(-\dfrac{3}{2}, 2\right)$ and the radius is $\sqrt{\dfrac{49}{4}} = \dfrac{7}{2}$ .

10. Graph the parabola $y = 3 + 2 x - x^2$ . Find the roots and the x and y-coordinates of the vertex.

The parabola opens downward.

$$\eqalign{ 3 + 2 x - x^2 & = 0 \cr x^2 - 2 x - 3 & = 0 \cr (x - 3)(x + 1) & = 0 \cr}$$

The roots are $x = -1$ and $x = 3$ .

The x-coordinate of the vertex is halfway between the roots: $x = \dfrac{-1 + 3}{2} = 1$ . The y-coordinate is

$$y = 3 + 2 \cdot 1 - 1^2 = 4.$$

Thus, the vertex is $(1, 4)$ .

$$\hbox{\epsfysize=1.5in \epsffile{rev4-1.eps}}\quad\halmos$$

11. Graph the parabola $y = x^2 - 4 x + 5$ . Find the roots and the x and y-coordinates of the vertex.

The parabola opens upward.

$$x = \dfrac{4 \pm \sqrt{16 - 20}}{2} = \dfrac{4 \pm 2 i}{2} = 2 \pm i.$$

The roots are $2 \pm i$ .

The x-coordinate of the vertex is $x = -\dfrac{b}{2 a} = 2$ . The y-coordinate is

$$y = 2^2 - 4 \cdot 2 + 5 = 1.$$

Thus, the vertex is $(2, 1)$ .

$$\hbox{\epsfysize=1.5in \epsffile{rev4-2.eps}}\quad\halmos$$

12. The area of a rectangle is 84 square miles. The length is 4 miles less than 3 times the width. Find the dimensions.

Let L be the length and let W be the width.

The area is 84 square miles: $84 = LW$ .

The length is 4 miles less than 3 times the width: $L = 3W - 4$ .

Substitute $L = 3W - 4$ into $84 = LW$ and multiply out:

$$84 = (3W - 4)W, \quad 84 = 3W^2 - 4W.$$

Solve for W:

$$\matrix{& 84 & = & 3W^2 & - & 4W & & \cr - & 84 & & & & & & 84\cr \noalign{\vskip2pt\hrule\vskip2pt} & 0 & = & 3W^2 & - & 4W & - & 84 \cr & 0 & = & & (3W + 14)(W - 6) & & & \cr & & & \swarrow & & \searrow & & \cr & & \matrix{& 3W & + & 14 & = & 0 \cr - & & & 14 & & 14 \cr \noalign{\vskip2pt\hrule\vskip2pt} & 3W & & & = & -14 \cr \div & 3 & & & & 3 \cr \noalign{\vskip2pt\hrule\vskip2pt} & W & & & = & -\dfrac{14}{3} \cr} & & & & \matrix{W - 6 & = & 0 \cr W & = & 6 \cr} & \cr}$$

$W = -\dfrac{14}{3}$ is ruled out, because the width of a rectangle can't be negative.

$W = 6$ gives $L = 3W - 4 = 3\cdot 6 - 4 = 14$ . The width is 6 miles and the length is 14 miles.

13. The length of a rectangle is 2 less than 3 times the width. The area is 176. Find the dimensions of the rectangle.

Let L be the length and let W be the width.

The area is 176, so $176 = LW$ .

The length is 2 less than 3 times the width: $L = 3W - 2$ .

Plug $L = 3W - 2$ into $176 = LW$ :

$$\eqalign{176 &= (3W - 2)W \cr 176 &= 3W^2 - 2W \cr 0 &= 3W^2 - 2W - 176 \cr}$$

Apply the quadratic formula:

$$W = \dfrac{2 \pm \sqrt{4 + 2112}}{6} = \dfrac{2 \pm \sqrt{2116}}{6} = \dfrac{2 \pm 46}{6} = \dfrac{-44}{6} \quad\hbox{or}\quad 8.$$

$W = \dfrac{-44}{6}$ doesn't make sense, because a width can't be negative. Therefore, the solution is $W = 8$ . The length is $L = 3W - 2 = 22$ .

14. The sum of two numbers is 5. The sum of their reciprocals is $\dfrac{45}{44}$ . Find the two numbers.

Let x and y be the two numbers.

The sum of two numbers is 5: $x + y = 5$ .

The sum of their reciprocals is $\dfrac{45}{44}$ : $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{45}{44}$ .

From $x + y = 5$ , I get $y = 5 - x$ . Plug this into $\dfrac{45}{44}$ : $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{45}{44}$ :

$$\dfrac{1}{x} + \dfrac{1}{5 - x} = \dfrac{45}{44}.$$

Clear denominators and simplify:

$$\eqalign{\dfrac{1}{x} + \dfrac{1}{5 - x} &= \dfrac{45}{44} \cr 44x(5 - x)\left(\dfrac{1}{x} + \dfrac{1}{5 - x}\right) &= 44x(5 - x)\cdot \dfrac{45}{44} \cr 44x(5 - x)\cdot\dfrac{1}{x} + 44x(5 - x)\cdot\dfrac{1}{5 - x} &= 44x(5 - x)\cdot \dfrac{45}{44} \cr 44(5 - x) + 44x &= 45x(5 - x) \cr 220 - 44x + 44x &= 225x - 45x^2 \cr 220 &= 225x - 45x^2 \cr 44 &= 45x - 9x^2 \cr 9x^2 - 45x + 44 &= 0 \cr}$$

Apply the quadratic formula:

$$x = \dfrac{45 \pm \sqrt{2025 - 1584}}{18} = \dfrac{45 \pm \sqrt{441}}{18} = \dfrac{45 \pm 21}{18}.$$

$x = \dfrac{45 + 21}{18} = \dfrac{66}{18} = \dfrac{11}{3}$ , which gives $y = 5 - \dfrac{11}{3} = \dfrac{4}{3}$ .

$x = \dfrac{45 - 21}{18} = \dfrac{24}{18} = \dfrac{4}{3}$ , which gives $y = 5 - \dfrac{4}{3} = \dfrac{11}{3}$ .

In either case, the two numbers are $\dfrac{4}{3}$ and $\dfrac{11}{3}$ .

15. Calvin and Bonzo, eating together, can eat 540 rib sandwiches in 6 hours. Eating alone, Calvin can eat 240 rib sandwiches in 4 hours less than it takes Bonzo, eating alone, to eat 240 rib sandwiches. How long does it take Calvin, eating alone, to eat 240 rib sandwiches?

Let x be Calvin's rate in sandwiches per hour, let y be Bonzo's rate in sandwiches per hour, and let t be the time it takes Calvin to eat 240 rib sandwiches.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & hours & & $\cdot$ & & sandiwches per hour & & $=$ & & sandwiches & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin and Bonzo & & 6 & & $\cdot$ & & $x + y$ & & $=$ & & 540 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin & & t & & $\cdot$ & & x & & $=$ & & 240 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bonzo & & $t + 4$ & & $\cdot$ & & y & & $=$ & & 240 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The second equation says $xt = 240$ , so $x = \dfrac{240}{t}$ .

The third equation says $y(t + 4) = 240$ , so $y = \dfrac{240}{t + 4}$ .

The first equation says

$$6(x + y) = 540, \quad\hbox{so}\quad x + y = 90.$$

Plug $x = \dfrac{240}{t}$ and $y = \dfrac{240}{t + 4}$ into $x + y = 90$ and solve for t:

$$\eqalign{ \dfrac{240}{t} + \dfrac{240}{t + 4} & = 90 \cr t(t + 4)\left(\dfrac{240}{t} + \dfrac{240}{t + 4}\right) & = 90t(t + 4) \cr 240(t + 4) + 240t & = 90t(t + 4) \cr 240t + 960 + 240t &= 90t^2 + 360t \cr 480t + 960 &= 90t^2 + 360t \cr 0 & = 90t^2 - 120t - 960 \cr 0 & = 3t^2 - 4t - 32 \cr}$$

(In the last step, I divided everything by 30.) Solve using the Quadratic Formula:

$$t = \dfrac{4 \pm \sqrt{16 + 384}}{6} = \dfrac{4 \pm \sqrt{400}}{6} = \dfrac{4 \pm 20}{6}.$$

Since t must be positive, $\dfrac{4 - 20}{6}$ is ruled out. The answer is

$$t = \dfrac{4 + 20}{6} = 4.\quad\halmos$$

16. A river flows at the rate of 8 feet per second. It takes Calvin 2 seconds longer to ride his boat 384 feet upstream against the current than it takes him to ride his boat 384 feet downstream with the current. Find the speed of Calvin's boat in still water (in feet per second).

Let x be the speed of Calvin's boat in still water. Let t be the time it takes him to ride 384 feet downstream.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & Time & & $\cdot$ & & Speed & & $=$ & & Distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Against current & & $t + 2$ & & $\cdot$ & & $x - 8$ & & $=$ & & 384 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & With current & & t & & $\cdot$ & & $x + 8$ & & $=$ & & 384 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The two rows give the equations

$$(t + 2)(x - 8) = 384 \quad\hbox{and}\quad t(x + 8) = 384.$$

From the second equation, $t = \dfrac{384}{x + 8}$ . Plug this into the first equation and solve for x:

$$\eqalign{ \left(\dfrac{384}{x + 8} + 2\right)(x - 8 & = 384 \cr \noalign{\vskip2pt} (x + 8) \cdot \left(\dfrac{384}{x + 8} + 2\right)(x - 8 & = (x + 8) \cdot 384 \cr \noalign{\vskip2pt} \left(384 + 2(x + 8)\right)(x - 8) & = 384(x + 8) \cr (2 x + 400)(x - 8) & = 384(x + 8) \cr 2 x^2 + 384 x - 3200 & = 384 x + 3072 \cr 2 x^2 - 3200 & = 3072 \cr 2 x^2 & = 6272 \cr x^2 & = 3136 \cr x & = \pm 56 \cr}$$

Since speed can't be negative, the speed of Calvin's boat is 56 feet per second.

17. Solve the inequality $x^2(x - 3)(x + 8) < 0$ .

$x^2(x - 3)(x + 8) = 0$ for $x = 0$ , $x = 3$ , and $x = -8$ .

$x^2(x - 3)(x + 8)$ is undefined for no values of x.

Plug in some test values and set up the sign chart.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & -9 & & -1 & & 1 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $x^2(x - 3)(x + 8)$ & & 972 & & -28 & & -18 & & 192 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=3in \epsffile{rev4-3.eps}}$$

The solution is $-8 < x < 0$ or $0 < x < 3$ ; in interval notation, it is $(-8, 0) \cup (0, 3)$ .

18. Solve the inequality $\dfrac{x^2}{x^2 - x - 6} > 0$ .

Write the inequality as

$$\dfrac{x^2}{(x - 3)(x + 2)} > 0.$$

$\dfrac{x^2}{(x - 3)(x + 2)} = 0$ for $x = 0$ .

$\dfrac{x^2}{(x - 3)(x + 2)}$ is undefined for $x = 3$ and $x = -2$ .

Plug in some test values and set up the sign chart.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & -3 & & -1 & & 1 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $\dfrac{x^2}{(x - 3)(x + 2)}$ & & $\dfrac{3}{2}$ & & $-\dfrac{1}{4}$ & & $-\dfrac{1}{6}$ & & $\dfrac{8}{3}$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfxsize=3in \epsffile{rev4-4.eps}}$$

The solution is $x < -3$ or $2 < x$ ; in interval notation, it is $(-\infty, -3) \cup (2, \infty)$ .

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