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(48) 2(b−c)y √2+a2=(b−c)2+ay3.

Ans. y=

√2(b−c)±√a3+(2-a) (b-c)2

a

184. If ba in the general form (x—a)(x—b)=0, it assumes the particular form (x-a)2=x2—2ax+a2=0.

If the two values of a bea and ―a, the form (x-a)(x+a)=x2—a2=0. 185. Recollecting that the value of the unknown quantity is called the root of the equation, it is seen that every equation of the second degree has two roots. and, by the general form (1), x2—(a+b)x+ab=0, that their sum is equal to the coefficient of the second term with the contrary sign, and that their product is equal to the absolute term or known quantity, when transposed to the first member. Thus, in Example 4, above, the sum of the two roots 3 and -9 is 6, and the product -27. The same may be seen in other examples.

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The general form ar+ bx c is capable of producing all the particular forms by the supposition of particular values for the coefficients. Thus, if b=0, it assumes the form of pure equations. If c=0, it may be written

x(ax+b)=0,

which we perceive may be verified by making x=0, or ax+b=0..x=

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b

a

Whenever an equation is

divisible throughout by the unknown quantity, one of its roots is zero.

When we know that the two roots of the equation of the second degree are real, the above relations make known at once the nature of these roots; for example, admitting that those of the equation x2-2x-7=0 are real, we conclude immediately that they are of different signs, because their product is equal to the absolute term -7, and, moreover, that the greater is positive, because their sum is +2, the coefficient of x taken with the contrary sign.

186. Another mode of solution may be derived as follows:

If we can, by any transformation, render the first member of the equation x2+px=q the perfect square of a binomial, a simple extraction of the square root will reduce the equation in question to a simple equation.

But (x+p) is x2+pr+p2.

In order, therefore, that the first member may be transformed to a perfect square, we must add to it the square of p; that is, the square of half the coefficient of the second term, or simple power of x; it thus becomes

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P

which is the square of x+. But since we have added

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member of the equation, in order that the equality between the two members may not be destroyed we must add the same quantity to the right-hand member also; the equation thus transformed will be

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the same form for the value of r as we obtained by the first method.

We affix the sign to

and also of

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-√+q, is +(22+q), and every quadratic equation must,

therefore, have two roots.

From what has just been said, we deduce the following general

RULE FOR THE SOLUTION OF A COMPLETE QUADRATIC EQUATION.

1. Transpose all the known quantities, when necessary, to one side of the equation, arrange all the terms involving the unknown quantity on the other side, and reduce the equation to the form ax2+bx=c.

2. Divide each side of the equation by the coefficient of x3.

3. Add to each side of the equation the square of half the coefficient of the simple power of x.

That member of the equation which involves the unknown quantity will thus be rendered a perfect square, and, extracting the root on both sides, the equation will be reduced to one of the first degree, which may be solved in the usual manner.

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Completing the square by adding to each side the square of half the coefficient of the second term,

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Either of these two numbers, when substituted for r in the original equation, will render the two members identical.

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The coefficient of x in this case is 1; .. in order to complete the square, we

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In the above example, the values of x contain imaginary quantities, and the roots of the equation are, therefore, said to be impossible.

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- a + b + x + 4 (a + b)2 = (a + b)2 + 4 (a+b)2

a+

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