Wherefore r=−2+12=10, or —2—12=-14. Where one of the values of x is positive and the other negative.

2. Given x2-12x+30=3, to find the value of x.
Here x2 - 12x=3−30=-27, by transposition,
Whence x=6+(36-27), by the rule,

Or, which is the same thing, x=6±√9,
Therefore x=6+3=9, or =6-3=3,
Where it appears that x has two positive values.
3. Given 2x2+8x-20=70, to find the value of x.
Here 2x2+8x=70+20=90, by transposition,
And x2+4x=45, by dividing by 2,

Whence x=-2+√(4+45), by the rule,
Or which is the same thing. x=−2±√49.
Therefore x=−2+7=5, or =—
:—2—7—9.

Where one of the values of x is positive and the other negative.

4. Given 3x2-3x+6=51, to find the value of x.

Whence x=

=±√(3), by the rule,

Or, by subtracting from

2

1

1

1

+

4'

36'

1

1

5. Given x2+20=423 to find the value of x.

by 2,

1

1

Here_x2 -_x=423–201=221 by transposition,

2

1

And x2-x=443, by dividing by or multiplied

2'

Whence we have x=¦±√(+443), by the rule,

1

Or, by adding and 44 together, x=

9

400

9

Therefore x=1+63=7, or=1—63=—6},

3

Where one value of x is positive and the other ne gative.

6. Given ax2 +bx=c, to find the value of x.

Here x2+x=by dividing each side by a.

α

b 1

·±√(b2+4ac).

2a 2a

7. Given ax2 bx+c=d, to find the value of x. Here ax2-bx=d-c, by transposition,

b d-c

--X--- by dividing by a.

α

And x2

α

Or, mults d -c & a by 4a, x= ·± —√(4α(d—c)+b2)

2a 2a

8. Given x+ax2=b, to find the value of x. Here x+ax2=b, by the question,

Whence x=±√(− = ± √(4b+a2)) by extraction of

10. Given 2x+3x=2, to find the value of x.

Here 2x3+3x=2, by the question,

And 23+x=1, by dividing by 2,

8'

Therefore x=(1)3=1, or (—2)3——8.

11. Given x-12x3+44x2-48x=9009(a), to find the

value of x.

2

2

4

Consequently, (x2-6x)2 +8 (x2-6x) = x2 −12x3 +44x2 —48x — a ; for since in extracting the square root of any quantity, the square of the root thus found plus the remainder is always equal to the proposed quantity. In a similar manner, the biquadratic equation x+2ax3 +5a2x2+4a3 d, may be exhibited under the form ខ

=

2

(x2+ax)*+4a3 (x* + ax)=d;

which can be resolved by the rule, page 130, for resolving quadratic equa

tions.

Hence it follows, that if the remainder, after having found the first two terms of the square root, according to the rule page 49, can be resolved into

Whence 22

-6x=−4±√(16+a), by the common rule, And, by a second operation, x=3+(9-4± √(16+a)) Therefore, by restoring the value of a, we have x=3±√(5±√9025)

Or, by extraction of roots, x=13, the Ans.

EXAMPLES FOR PRACTICE.*

1. Given x2 -8x+10=19, to find the value of x.

Ans. x9.

2. Given x2 -x-40=170, to find the value of x.

Ans x=15.

3. Given 3x2+2x-9=76, to find the value of x.

1

1

Ans. x=5.

4. Given x2+73=8, to find the value of x.

1 1
31

Ans. x=1.

5. Given x—√x=22}, to find the value of x.

ལ་

Ans. x=49.

6.† Given x+(5x+10)=8, to find the value of x

Ans. x=3.

two such factors, so that the factor containing the unknown quantity, shal be equal to the terms of the root thus found; the proposed biquadratic may be always reduced to a quadratic form, as above. See Ryan's Algebra page

396.

ED.

*The unknown quantity in each of the following examples, as well as in *those given above, has always two values, as appears from the common rule; but the negative and imaginary roots being, in general, but seldom used in practical questions of this kind, are here suppressed.

+ In some quadratic equations involving radical quantities of the form✔ (ax+b), both the values of x, found by the ordinary process, will not answer the proposed equation, except we take the radical quantity with the double sign. In resolving the above example, two values of x, that is, 18 and 3, are found; but, it appears, that 18 does not answer the condition of the equa tion except we take the radical quantity (5x+10) with the sine.

Now, since these two values of x are formed from the resolution of the equation x2--21x=-54; it necessarily follows that each of them, when substituted for x, must satisfy that equation; which may be verified thus; in the first place, by substituting 18 for x, in the equation x2-21x-54, we have (18)2-11 X 18-54, or 324-378 -54, that is, -64 —— -54, or by transposition 0= 0.

Again, substituting 3 for x, we have (3)2--21X354, or 9-63—54; ..54-54-0, or 0~0.

And as the equation x2-21x=-54, may be deducted from the equation +(5x+10)—8—x, or — ✔ (5x+10)=8-x; it is evident that the radical quantity ✔(5x+10) must be taken with the double sign±, in the pri

7. Given (10+x) −/(10+x)=2, to find the value of x.

Ans. x=6.

8. Given 2x4 -x2+96—99, to find the value of x. 1 Ans. x=√6.

9. Given x+20x3-10=59, to find the value of x. Ans. x=2/3.

10. Given 3x2n−2x2+3=11, to find the value of x. Ans. x=2.

2

1 2

12. Given (3+2x)=+x2, to find the value of

14. Given =√(1 − x3)=x2, to find the value of x.

x

15. Given x√(−1)=√(x2-b2), to find the valuc

of x.

1

Ans. x=‡a+¦ ¦√(8b2+a2).

1

16. Given ✓/(1+x-x2) — 2(1+x-x2)= to find the

value of x.

1 1 Ans. x=6 +41.

mitive equation, in order that it would be satisfied by the values, 18 and 3 of x, found above; that is, 18 answers to the sign and 3 to the sign+ See Ryan's Elementary Treatise on Algebra, Theoretical and Practical where this subject is clearly illustrated.

En.