Hence

a + B − (y + d) = √1+4×132=√529=23;

Let x=y+z; then y3 +23+ (3yz − 9) (y + z) — 14 =0. Also let 332-9=0, then y3 +23 = 14 ;

: 3 - 3 = VIA - 4 x 3 = ± 2 V22.

Hence y3=7±√22, 23=7+ √22;

Since

applicable.

8213

(1213)* – (52)* is negative, Cardan's method is not

-

Hence, by the tables,

30=35° 14' 7"187; and .. = 11° 44′ 42′′• 396 ;

··. y = 2 √/82 cos 0 = 5'9105278.

3

2

Also y = √52 cos(2+3), or 3 √82 cos (230)

Let (x2+ex+f) (x2 — ex + g) = x4 — 3x2 - 42x — 40 = 0.

Equating the coefficients of the same powers of x, we get g+f-e2=-3, eg-ef=-42, fg = -40.

Then

Let ey, then y3 — бy2 + 169y — 1764 =0.

By trial of square numbers, 9 is found to be a root of this cubic;

.. e = 3, ƒ= 1 (9 − 3 + 14) = 10, g=−4.

2.

x-25x60x-36 = 0.

By Euler's Method.

Let x=y+z+u.

Then x2=y2+z2 + u2 + 2 (yz + yu+zu) ;

·· x1 — 2x2 (y2 + z2 + u3) + (y2 + z2 + u3)3

= 4 (y3z3 + y3u3 + z3u3) + 8yzu (y + z + u),

oг. x1 — 2x2 (y2 + z2 + u3) — 8xyzu + (y2+z2 + u3)2

Comparing this equation with the proposed

x1 — 25×3 — 60x - 360, we have

y3 + z3 + u2 = 25, yzu =

2

бо

= 15

8

(y3 + z3 + u2)3 − 4 (y3z3 + y3u2 + z1u3) = − 36 ;

Hence the values of y', z', u' are the roots of the equation

Let t=2;

Let t; then k3 - 50k2 + 769k - 3600 = 0, the roots of

4

which are 9, 16, 25; .. the values of t are

Now x=y+z+u, with the condition that yzu =+ 15;

Take away the second term by putting xy + 1, the result is

y1 — 6y3 — 16y +21=0.

Then y+6y2+ 16y — 21,

=

add 2ky+k to both sides;

•. (y2+ k)2 = (2k +6) y2 + 16y + k2 - 21 a perfect square;

k3 +3k2 - 21k-950; of which 5 is a root.

Hence y2+5+ (4y+ 2);

=

.. y2-4y+3=0;.. y = 1, and y=3,

y2+4y+7=0; . y=-2±√3.

The roots of the original are.. 4, 2, 1± √−3.

LIMITS OF ROOTS.

Ex. 14.

1. Transform the equation into one whose roots shall be less by h than the roots of the given equation,

then (y+h)3-4 (y + h) 3 − 4 (y + h) + 20 = 0,

or (h3 — 4h3 — 4h +20) + (3h3 — 8h − 4) y+(3h−4) y2+y3=0...(1).

Substitute the numbers 1, 2, 3, 4, &c. successively for h in (1), and the first number which makes all the coefficients positive will be found to be 4;

.. 4 is the No. next greater than the greatest positive root.

And substituting the numbers 1, 2, 3, 4, &c. successively for x in these expressions, we find that 4 is the first number that makes them all positive;

.4 is the superior limit to the roots.

10. There are but two changes of sign, viz. from + to -, and from to +, therefore there cannot be more than 2 positive roots.

Again, there is but one permanence of the same sign, viz. -,-, therefore the equation cannot have more than I negative

root.

RATIONAL ROOTS.

Ex. 15.

1. Here the limits of the roots are 7 and divisors of the last term are therefore 6, 4, 3, 2;

1, and the

therefore 6 is the only commensurable root, since it does not satisfy the equation 3x2-18x+22=0; .. dividing the given equation by x-6, we have

4.

x2-3x+4=0,

whence x = = {3 ± (−7)*}.

x3- 5x2-18x+72=0= f(x) suppose.

Changing the signs of the alternate terms, we have