4. In the equation z2 2cx=a2 to find the two values of r.

Solu. Add c2 to each side, then r2

• 2cx+c2=a2+c2.

Now extract the root, and x-c= ±√ a2+c2 ;

Therefore, x=c+√/a2+c2, or c

√ a2+c2.

Examples for Practice.

5. Let x2-2x=15, to find the two values of x... Ans. x=5 or-3.

(73). Every equation, in which the unknown quantity is found in only two terms, and the index of the highest power is double the index of the lowest, may be reduced to the form of a quadratic equation, and solved according to the foregoing rules.

(74). For Example. Let x2+Px"=Q be any general equa

tion.

Solu. Let x=y, then ry;

substitute now these values for 2" and ", and the given equation becomes y2+Py=Q.

1. Let x2-6x=27, to find the values of x in this equation.

*=y,}

Solu. Let x=y, whence the equation becomes y2-бy=27 ; then x*=y2;

Therefore y-3+ √36=+6, and y=+6+3=9 or -3. But, because 2=y, x=±√ÿ=±√9, or ±√—3=+3, or + √-3.

2. Find the values of x in the equation x-2x3=48.

Solu. Let ry, therefore, by substituting y2-2y=48. then x=y2;

But y-1+49=±7; hence, y=+7+1=8 or-6; consequently, x=✔y=8, or V-6=2, or 3—6.

Examples for Practice

3. Find the values, of x in the equation x+4x=12

Ans. x=√2, or + √6

4. Find the values of x in the equation x6-8x=513.

Ans. x=3 or 3—19.

5. Find the values of x in the equation 1”—4x”—10.

7

Solu. First, multiply by x+1, and the equation;

2 + =5, x+1 x

=

3. Let 5x-90x3-270=945, to find in this equation the values

of x.

Solu. By transposition,

5x-90r-270=945=5x-90x=945+270=1215.

Then divide by 5, and we get x-18x3=243.

Complete the square, and it becomes

x—18x3+81=243+81=324.

Extract now the square root, and 2—9=±√/324=+18.
Therefore +18+9=27, or-9;

And r2=27, or 4-9-3, or -9.

18x3 =

243, as

1

Nole.-Without the formality of substituting y "for x6, and y for x3, we
proceeded at once to the solution of the equation, 26.
exemplified in the second step of the operation.

4. Find, in the equation 3ax2-2b=4x-5c, the values of x.
Solu. First, by transposition, we get 3ax2-4x=2b—5c.

5. Let x2-7x+1=9, to find the values of x.

Ans. x=8, or — 1.

6. Given 4x2-7x=492, required the values of x.

Ans. x=12, or — - 10

7. Let x2-6x+19=13, to find r's values.

Ans. x=4.732, or 1.268.

8. Given 5x2+4x=25, required the values of r.

Ans. x=1.871, or —

9. Let 22—

*—1=x+11, to find the values of x.

6

Ans. x=12, or — 6.

10 Given 1+1=1,
+-—=—, required the values of x.

– 2.671.

Ans. x=3, or

E

15. Given 2x*-7x2=9, to find the values of x.

13

to find the values of x.

Ans. x=2, or - 3.

Ans. x+3, or ± √

16. Let 2x1-r2=496, required the values of x.

· Ans. x=±4, or ±√ —31

17. Given x-ra to find the values of x.

2

Ans. x=3

1+√4a+1 2

18. Let 2x+15=3r required the values of x.

20. Let 4ax-26x=c, required the values of x.

75) PROBLEMS, or questions, involving quadratic equations, are solved by the same process as that for the solution of those which contain simple equations. But as every quadratic

equation contains two values of the unknown quantity; ques tions of this class may admit of a double solution. And as the particular nature of the question must determine whether one or both of these values apply to it, no general rule can be laid down; yet may the processes of solution of Problems involving quadratic equations be sufficiently understood by observing the conclusions deduced from the following examples.

EXAM. I. To find that number to which if you add 12, and multiply the sum by the number required, the product shall be 589.

Solu. Let r = the number sought;

Then (r+12) x =589, or x2 + 12r 589 by the question. Complete the square and we have x2+12x+36 = 625; hence, Extract the root, we get +6 = √625

€ 25

Therefore, x = 25—6— 19, the number sought.

EXAM. II. To divide the number 56 into two such parts, that their product shall be 640.

Solu. Let x = one part; then 56

the other part;

x2 640 by the

And xx (56 - x) =the product of the two parts.
Hence, x x (56 — x) =640, or 56x

question, or x2 56x=

640; hence

By completing the square, 256x+784784640 144;

Therefore, x

-

28= ± √144=

and, consequently, x = 28 + 12 = 40 or 16.

12;

Note. The two values of the unknown quantity are, in this instance, th two parts into which the given number was required to be divided.

EXAM. III. Let the difference of two numbers be 7, and half their product plus 30 be equal to the square of the less number; it is required to find those two numbers: