Chapter 2

Exer. 1-50: Solve the equation. $$ 36 x^{-4}-13 x^{-2}+1=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}-6 x+13=0 $$

Exer. \(31-44\) : Solve by using the quadratic formula. $$ \frac{5}{w^{2}}-\frac{10}{w}+2=0 $$

Archeologists can determine the height of a human without having a complete skeleton. If an archeologist finds only a humerus, then the height of the individual can be determined by using a simple linear relationship. (The humerus is the bone between the shoulder and the elbow.) For a female, if \(x\) is the length of the humerus (in centimeters), then her height \(h\) (in centimeters) can be determined using the formula \(h=65+3.14 x\). For a male, \(h=73.6+3.0 x\) should be used. (a) A female skeleton having a 30 -centimeter humerus is found. Find the woman's height at death. (b) A person's height will typically decrease by \(0.06\) centimeter each year after age 30 . A complete male skeleton is found. The humerus is 34 centimeters, and the man's height was 174 centimeters. Determine his approximate age at death.

Solve the equation. $$\frac{2}{2 x+5}+\frac{3}{2 x-5}=\frac{10 x+5}{4 x^{2}-25}$$

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ x^{4} \geq x^{2} $$

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ 2 x(6 x+5)<(3 x-2)(4 x+1) $$

Exer. 1-50: Solve the equation. $$ x^{-2}-2 x^{-1}-35=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}-2 x+26=0 $$

Exer. \(31-44\) : Solve by using the quadratic formula. $$ \frac{x+1}{3 x+2}=\frac{x-2}{2 x-3} $$

Solve the equation. $$\frac{2}{2 x+1}-\frac{3}{2 x-1}=\frac{-2 x+7}{4 x^{2}-1}$$

Exer. 41-42: As a particle moves along a straight path, its speed \(v\) (in \(\mathrm{cm} / \mathrm{sec}\) ) at time \(t\) (in seconds) is given by the equation. For what subintervals of the given time interval \([a, b]\) will its speed be at least \(k \mathrm{~cm} / \mathrm{sec}\) ? $$ v=t^{3}-3 t^{2}-4 t+20 ; \quad[0,5] ; \quad k=8 $$

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{4}{3 x+2} \geq 0 $$

Exer. 1-50: Solve the equation. $$ 3 x^{2 / 3}+4 x^{1 / 3}-4=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}+4 x+13=0 $$

Exer. \(31-44\) : Solve by using the quadratic formula. $$ 4 x^{2}+81=36 x $$

Solve the equation. $$\frac{3}{2 x+5}+\frac{4}{2 x-5}=\frac{14 x+3}{4 x^{2}-25}$$

Exer. 41-42: As a particle moves along a straight path, its speed \(v\) (in \(\mathrm{cm} / \mathrm{sec}\) ) at time \(t\) (in seconds) is given by the equation. For what subintervals of the given time interval \([a, b]\) will its speed be at least \(k \mathrm{~cm} / \mathrm{sec}\) ? $$ v=t^{4}-4 t^{2}+10 ; \quad[1,6] ; \quad k=10 $$

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{3}{2 x+5} \leq 0 $$

Exer. 1-50: Solve the equation. $$ 2 y^{1 / 3}-3 y^{1 / 6}+1=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}+8 x+17=0 $$

Exer. \(31-44\) : Solve by using the quadratic formula. $$ 24 x+9=-16 x^{2} $$

Solve the equation. $$\frac{5}{2 x+3}+\frac{4}{2 x-3}=\frac{14 x+3}{4 x^{2}-9}$$

Guinness Book of World Records reports that German shepherds can make vertical leaps of over 10 feet when scaling walls. If the distance \(s\) (in feet) off the ground after \(t\) seconds is given by the equation \(s=-16 t^{2}+24 t+1\), for how many seconds is the dog more than 9 feet off the ground?

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{-2}{4-3 x}>0 $$

Exer. 1-50: Solve the equation. $$ 6 w+7 w^{1 / 2}-20=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}-5 x+20=0 $$

Exer. \(31-44\) : Solve by using the quadratic formula. $$ \frac{5 x}{x^{2}+9}=-1 $$

Solve the equation. $$\frac{-3}{x+4}+\frac{7}{x-4}=\frac{-5 x+4}{x^{2}-16}$$

If an object is projected vertically upward from ground level with an initial velocity of \(320 \mathrm{ft} / \mathrm{sec}\), then its distance \(s\) above the ground after \(t\) seconds is given by \(s=-16 t^{2}+320 t\). For what values of \(t\) will the object be more than 1536 feet above the ground?

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{-3}{2-x}<0 $$

Exer. 1-50: Solve the equation. $$ 8 t-22 t^{1 / 2}-21=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{2}+3 x+6=0 $$

Exer. \(31-44\) : Solve by using the quadratic formula. $$ \frac{1}{7} x^{2}+1=\frac{4}{7} x $$

Show that the equation is an identity. $$(4 x-3)^{2}-16 x^{2}=9-24 x$$

The braking distance \(d\) (in feet) of a certain car traveling \(v \mathrm{mi} / \mathrm{hr}\) is given by the equation \(d=v+\left(v^{2} / 20\right)\). Determine the velocities that result in braking distances of less than 75 feet.

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{2}{(1-x)^{2}}>0 $$

Exer. 1-50: Solve the equation. $$ 2 x^{-2 / 3}-7 x^{-1 / 3}-15=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ 4 x^{2}+x+3=0 $$

Exer. 45-48: Use the quadratic formula to factor the expressions. $$ x^{2}+x-30 $$

Show that the equation is an identity. $$(3 x-4)(2 x+1)+5 x=6 x^{2}-4$$

The number of miles \(M\) that a certain compact car can travel on 1 gallon of gasoline is related to its speed \(v\) (in mi/hr) by $$ M=-\frac{1}{30} v^{2}+\frac{5}{2} v \text { for } 0

Exer. 1-50: Solve the equation. $$ 6 u^{-1 / 2}-13 u^{-1 / 4}+6=0 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ -3 x^{2}+x-5=0 $$

Show that the equation is an identity. $$\frac{x^{2}-9}{x+3}=x-3$$

For a particular salmon population, the relationship between the number \(S\) of spawners and the number \(R\) of offspring that survive to maturity is given by the formula \(R=4500 S /(S+500)\). Under what conditions is \(R>S\) ?

Exer. 1-50: Solve the equation. $$ \left(\frac{t}{t+1}\right)^{2}-\frac{2 t}{t+1}-8=0 $$

Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ |x|<3 $$

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{3}+125=0 $$

Show that the equation is an identity. $$\frac{x^{3}+8}{x+2}=x^{2}-2 x+4$$