17

15

tracting the root, x+==+, and by transposition, x——8,

or-1.

Ex. 5. Given 4x2-3x=85, to find the values of x.

(Art. 417). Multiplying by 16, 64x2-48x=1360, and, adding the square of 3, 64x2-48x+9=1369;

... extracting the square root, 8x-3=37;

by transposition, 8x=40, or −34, .'.x=5, or—41.

of x.

Multiplying by 2x-6, we have 10x2-36x+6=4x2 – 12x +3x-15x+18;

.. by transposition, 3x2-9x=12;

and by division, x2-3x=4 :

.. completing the square, 2-3x+2=4+1=2, and extracting the root, x-3=;

... 63x-14x2+6x2—23x+10=378-84x ;

by transposition, 124x-8x2-368,

and x

—31x——46 ; .'. by completing the square,

31 961 961
-X +

225

x2

-46- ;

16 16

16

Ex. 9. Given

Dividing by

2'

x3+√x3=6√x, to find the values of x. x, x2+x=6 :

... completing the square, x2+x+1=6+1= ; and extracting the root, x++;

.*. x=2, or-3.

n

Ex. 10. Given x2-2ax2=b, to find the values of x.

n

"—2ax+a2=a2+b;

Completing the square, x-2ax7.

n

... extracting the root, x-a=±√(a2+b),

n

and x =a±√(a2+b) ; .•. x=(a±√(a2+b))".

Ex. 11. Given x2—2x+61√√(x2—2x+5)=11, to find the values of x.

Adding 5 to each side of the equation,

(x2-2x+5)+6/(x2-2x+5)=16;

... by completing the square,

(x2-2x+5)+6✓/(x2-2x+5)+9=25;

and extracting the root,

(x2-2x+5)+3=±5;

..√(x2-2x+5)=2, or -8;

.. squaring both sides, x2-2x+5=4, or 64 ; whence x2-2x+1=0, or 60 ;

and extracting the root, x-1=0, or ±√60; ... x=1, or 1±√60.

SCHOL. It is proper to observe, that the equation, x2-2x +1, has two equal roots, although x appears to have only one value; but it is because x is twice found 1, as the common method of resolution shows; for we have x=1±√✓0, that is to say, is in two ways=1.

Ex. 12. Given x2+4x3+12x2+16x=a, to find the values of x.

Here the two first terms of the square root of the left-hand member (Art. 299), is found to be x2+2x, and the remainder is 8x+16x which can be readily resolved into the factors 8 and x2+2x, since (8x2+16x)(x2+2x) gives 8 for the quotient. Consequently the proposed equation may be exhibited under the quadratic form (x2+2x)2+8(x2+2x)=a; ... by completing the square, (x2+2x)2+8(x2+2x)+16=a+ 16; and extracting the root, x+2x+4=±√(a+16) Now by taking the positive sign,

x2+2x+4=+√(a+16);

by transposition, x2+2x=−4+√(a+16);

... completing the square, x+2x+1=−3+√(a+16) ;

and extracting the root, x+1=±√(−3+√(a+16));

••.x=-1±√(-3+√(a+16)).

Again, by taking the negative sign.

x2+2x+4=-√(a+16) ; ·

x2+2x=-4-(a+16); and

completing the square, x+2x+1=—3—√(a+16); ... extracting the root, x+1=±√(−3−√(a+16)); and x=-1+(-3-√(a+16)).

Ex. 13. Given 3x2-2x+12=16-4, to find the values

of x.

By transposition, 3x2-12x=16-4-12=0;

and by division, x2-4x=0; .. by completing the square, x2-4x+4=4, and extracting the root, x-2=+2 ; ..=4, or 0. See (Art. 405).

Ex. 14. Given x3-4x2+6x=4, to find the values of x. (Art. 423), multiplying both sides by x, x-4x+6x2-4x

=0,

(Art. 422)..(x2-2x)2+2(x2-2x)=0. ..x2—2x+1=+1, and x=1±√±1; ..the three roots of the proposed equation, are 1, 1+√−1, and 1-1. The other value of x, which is equal to 1-1, or 0, belongs to the equation (x2-2x)2+2(x2-2x)=0; hence there are four roots, or four values of x, which will satisfy this last equation.

841 17 232 1

Ex. 15. Given 27x2 - -+·
3x2 3 3x

values of x.

Multiplying every term by 3,

3x2+5, to find the

841
81x2+17=
x2

232 1
x2

+15;

1

... by transposition, 81x2+17+.

841 232
+ +15.

x2 x2 τ

Adding unity to each side, in order to complete the square;

31z2+18+

1 841 232

+: +16;

x2 x2 x

and extracting the root, 9x+==±(—+4).

x

Let the positive value be taken; then by transposition, 9x 4. and.. 9x-4x=28; by completing the square, &c.,

taken, 9x2+4x=-30 and completing the square, &c., x= −2+√(−266)

9

Ex. 16. Given 3x2+2x—9—76, to find the values of x.

Ans. x=4, or -21.

Ex. 20. Given √(x+5)×√(x+12)=12, to find the values of x. Ex. 21. Given 2x2+3x—5√(2x2+3x+9)+3=0, to find 9 -3+/-55 or 2'

Ans. x=3, or

4

the values of x. Ex. 22. Given 9x+√(16x2+36x3)=15x2-4, to find the

4

Ans. x=
x=3' or --

1

9±√481

values of x.

50

Ex. 24. Given x2-2x3-x=132, to find the values of x.

3

Ex. 25. Given x5+x3756, to find the values of x.

Ans. x=243, or (-28).

Ex. 26. Given x3-x=56, to find the values of x.

Ans. x=4, or (—7)3.

Ex. 27. Given x+5=√(x+5)+6, to find the values of x. Ans. x=4, or — 1.

Ex. 28. Given x+16-7(+16)=10-4/(x+16), to

Ans. x=9, or

-12.

=13, to find the values of x. -2.

Ans. x=4, or

Ans. x=4, or—43.

Ex. 34. Given √(4x+5)×√(7x+1)=30, to find the va

lues of x.

Ans. x=5, or—61}.

to find the values of x.

Ans. x=3, or

-15.

x x+12 15'

Ex. 36. Given x3+7x=44, to find the values of x.
Ans. x=±8, or ±(−11)*.

Ex. 37. Given 4x+x=39, to find the values of x.

Ans. x=729, or

Ex. 38. Given 3x+42x3=3321, to find the values of x.

Ans. x=3, or —3/41.

to find the values of x.

Ans. x=4, or 1/2.

Ex. 40. Given x2+11+√(x2+11)=42, to find the values Áns. x+5, or ±√38.

of x. Ex. 41. Given x2-12x+50=0, to find the values of x. Ans. x=6+(-14).

Ex. 42. Given 3x-x2=10, to find the values of x.

Ans. x=6+-4. Ex. 43. Given xo—2x3—48, to find the values of x.

Ans. x=2, or 3-6.

Ex. 44. Given x2+2x3-7x2-8x=-12, to find the values

of x.

Ex. 45. Given

values of x.

Ans. 2, or 3, or 1, or —2. x2-10x3+35x2-50x+24=0, to find the Ans x=1, 2, 3, or 4.

Ex. 46. Given x3-8x2+19x-12=0, to find the values

of x.

Ans. x=1, 3, or 4.