EXERCISES

Solve and check your answers in the following. If you meet with a surd, find its approximate value by use of the tables.

EXERCISES - APPLIED PROBLEMS

If surds occur in the answers, get their approximate values by means of the tables.

1. If three times the square of a certain number be diminished by 52, the result is 20 more than the square of the number itself. What is the number?

2. One side of a certain right triangle is 9 inches long, and the hypotenuse is 15 inches long. How long is the other side?

[HINT. Work by algebra, using the principle in Ex. 20, p. 21.]

3. How far apart are the two diagonally opposite corners of a room that is 12 feet wide and 15 feet long?

4. What is the length of the longest umbrella that can be placed in the bottom of a trunk the inside of which is 32 inches by 20 inches?

5. A square field contains 5 acres. How many rods of fence will be required to inclose it?

6. The sides of a 90-acre field are to each other as 2 is to 4. Find the length and breadth.

The area of

[HINT. Let x= = the length and y=the width. Then we must have x/y=(2)/4. Solving this for y gives y=x. the field, or xy, thus becomes x2. Now find x.]

7. Find the mean proportional between 27 and 12.

[HINT. Let x be the number desired. Then we must have 27/x=x/12. See § 133.]

8. Find the mean proportional between 16 and 21. State any essential difference between your answer and that obtained for Ex. 7.

9. In the semicircle ABC the line PQ is perpendicular to the base and divides it into parts which are 5 inches long and 8 inches long respectively. How long is PQ?

[HINT. See Ex. 8, p. 197.]

10. From a certain point P outside a circle a secant is drawn cutting the circle at the points Q and R. If PQ is 6 inches and PR is 11 inches how long is the tangent PT?

[HINT. See Ex. 10, p. 197.]

ΑΙ

R

Q B

5

P

8

FIG. 78.

FIG. 79.

F

D

A

E

FIG. 80.

B

C

11. If you have a board 20 inches square how wide a strip must be taken from all sides so that the remaining square shall have one half the area of the original piece? 5(2√2) in. Ans.

12. We know that a window whose lower part is a rectangle and whose upper part is a semicircle admits the most light when the width and height are the same. If the area of such a window is 414 square feet, what is the width?

[HINT. Take π=34.]

13. An 8-inch square was taken from each corner of a square piece of tin. The sides were then turned up so as to form a box containing 1152 cubic inches. How long was the piece of tin?

14. What is the formula for the length of the tangent from a point to a circle if the secant from that point is divided by the circle into segments (parts) equal respectively to a and b?

15. What is the formula for the radius of the circle whose area is A? Apply your answer to find the radius in case the area is 37 square inches?

16. One of the sides of a certain triangle is a units long. What is the formula for the corresponding side of a triangle which is similar to the first one, but whose area is b times as great?

[HINT. See § 115.]

17. Find the formula for the radius of the sphere whose surface is n times as great as that of the sphere whose radius

is r.

PART II. SOLUTION OF AFFECTED QUADRATICS

153. Solution of Affected Quadratics by Factoring. An affected quadratic may often be solved by factoring. This was explained in § 68, which the pupil should now read again, but the following examples will further illustrate the method.

EXERCISES

Solve by factoring and check your answers for Exs. 1-10. 1. x2+8x-48=0.

SOLUTION. x2+8 x−48= (x−4)(x+12). (See § 58.)

The desired solutions are therefore those obtained by solving the simple equations x-4=0 and x+12=0. (See § 68.)

Hence, the solutions are x=4 and x=-12. Ans.

CHECK. Substituting 4 for x in the given equation gives 42+8x4-48, which reduces to 16+32-48, and this =0 as required. Similarly, if we substitute - 12 for x we obtain (−12)2+8×(−12)–48 which 144-96-48-0, as required.

2. x2-x-12=0.

3. x2-12x+35=0.

4. x2-7x=8.

[HINT. Remember to transpose the 8 so as to have the right-hand member zero. See § 68.]

15. (x+4)(x-2)=11(x-2).

[HINT. Write as (x-2)[(x+4)-11]=0 and apply the principle

154. Solution by Completing the Square. We often meet with quadratics, such as

x2+7x-5=0,

which we cannot solve by factoring, as in § 153. The difficulty here is that we cannot factor x2+7x-5. However, this quadratic and all others (whether they be solvable by factoring or not) can be solved by a process called completing the square. How this is done will be best understood from a careful study of the following examples.

NOTE. At this point the pupil should review § 58.

EXAMPLE 1. Solve the equation

x2+6x=16.

SOLUTION. The first member of this equation, or x2+6 x, would become a perfect trinomial square if 9 were added to it (see § 58). Our first step, therefore, is to add 9 to both sides of the equation, thus "completing the square' on the left side and at the same time not destroying the equality of the two members. This gives the equation

x2+6x+9=25,

or,

(x+3)2=25.