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but from the preceding principle, this equation can only be true when x a = 0, or a = a; making this supposition, and dividing both members of the given equation by ✓–

1, we have b = y, which was to be shown.

In the same way, it may be shown that, when a + b = x + Vy, we have, a = x, and b=y: that is, in all equations of this form, the rational and radical parts, in each member, are respectively equal to each other.

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3o. The product of two factors, of the forms,

(a + b V-1), and x (a 6V1), is positive for all real values of x:

For, performing the multiplication, we find the product equal to,

22 2ax + a + b?,

which can be written under the form,

(x – a)? + 12.

Now, whatever may be the value of x, the part (2 - a)? will be positive, since it is a square; 62 is also positive; hence, their sum, or the required product, is also positive, which was to be proved.

III. SOLUTION OF RADICAL EQUATIONS.

139. Radical equations are equations containing radical quantities. No fixed rules can be given for solving such equations. The methods of proceeding will be best illustrated by examples.

EXAMPLES.

1. Given Vã + 16 = 2 + , to find X.

Squaring,

X + 16

= 4 + 4V + x; transposing and reducing,

Væ = 3; squaring,

X = 9.

2. Given 1 - V1 - x = n(1 + V1 — x), to find x. Reducing,

1 - n = (n + 1) V1 -2; squaring,

1 – 2n + n2 = (n2 + 2n + 1) (1 — 2); whence,

4n
(1 + n)a.

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3. Given Va + 2 va - Nax, to find x. Squaring and reducing,

2a ax = 2 Va? – X?, squaring again,

4a2 4a2x + a x2 = 4a? reducing, and dividing by x,

(a? + 4)X = 4ao; whence,

4a2
a? + 4

- 4oc? ;

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;

Transposing,

81(3 + 431) 16x - 8/3/x + 3 =

(4Væ- 13) (4Væ+ V3) factoring,

81 V3(V3 + 4Væ) (47x-V3)2

(4Væ– 73)(4Væ+ 1/3)

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dividing by 4, and squaring,

(31/3 + 13)

= 3. 16

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72 - 2ab 23. a+x+va? + 6x + x = b.

Ans. X =

36 2a*

4(x—16) 24.

Ans. x = 564 Vx+3 VX-2 Væ+4.

X-9

X-4

+

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